Chapter03rw-final

Chapter03rw-final - 3 3.1 3.2 Stoichiometry: Calculations...

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Unformatted text preview: 3 3.1 3.2 Stoichiometry: Calculations with Chemical Formulas and Equations Visualizing Concepts Reactant A = red, reactant B = blue Overall, 4 red A 2 molecules + 4 blue B atoms → 4 A 2 B molecules Since 4 is a common factor, this equation reduces to equation (a). Write the balanced equation for the reaction. 2H 2 + CO → CH 3 OH The combining ratio of H 2 : CO is 2:1. If we have 8 H 2 molecules, 4 CO molecules are required for complete reaction. Alternatively, you could examine the atom ratios in the formula of CH 3 OH, but the balanced equation is most direct. 3.3 (a) (b) There are twice as many O atoms as N atoms, so the empirical formula of the original compound is NO 2 . No, because we have no way of knowing whether the empirical and molecular formulas are the same. NO 2 represents the simplest ratio of atoms in a molecule but not the only possible molecular formula. 3.4 The box contains 4 C atoms and 16 H atoms, so the empirical formula of the hydrocarbon is CH 4 . (a) Analyze. Given the molecular model, write the molecular formula. Plan. Use the colors of the atoms (spheres) in the model to determine the number of atoms of each element. Solve. Observe 2 gray C atoms, 5 white H atoms, 1 blue N atom, 2 red O atoms. C 2 H 5 NO 2 3.5 38 3 Stoichiometry (b) 2 C atoms = 2(12.0 amu) = 24.0 amu 5 H atoms = 5(1.0 amu) = 5.0 amu 1 N atoms = 1(14.0 amu) = 14.0 amu 2 O atoms = 2(16.0 amu) = 32.0 amu 75.0 amu Solutions to Exercises Plan. Follow the method in Sample Exercise 3.9. Calculate formula weight in amu and molar mass in grams. Formula weight = 75.0 amu, molar mass = 75.0 g/mol (c) Plan. The molar mass of a substance provides the factor for converting moles to grams (or grams to moles). Solve. 3 mol glycine × (d) 75.0 g glycine mol = 225 g glycine Plan. Use the definition of mass % and the results from parts (a) and (b) above to find mass % N in glycine. Solve. mass % N = gN g C 2 H 5 NO 2 × 100 Assume 1 mol C 2 H 5 NO 2 . From the molecular formula of glycine [part (a)], there is 1 mol N/mol glycine. mass % N = 14.0 g 1 × (molar mass N) × 100 = × 100 = 18.7% molar mass glycine 75.0 g 3.6 Analyze. Given: 4.0 mol CH 4 . Find: mol CO and mol H 2 Plan. Examine the boxes to determine the CH 4 :CO mol ratio and CH 4 :H 2 O mole ratio. Solve. There are 2 CH 4 molecules in the reactant box and 2 CO molecules in the product box. The mole ratio is 2:2 or 1:1. Therefore, 4.0 mol CH 4 can produce 4.0 mol CO. There are 2 CH 4 molecules in the reactant box and 6 H 2 molecules in the product box. The mole ratio is 2:6 or 1:3. So, 4.0 mol CH 4 can produce 12:0 mol H 2 . Check. Use proportions. 2 mol CH 4 /2 mol CO = 4 mol CH 4 /4 mol CO; 2 mol CH 4 /6 mol H 2 = 4 mol CH 4 /12 mol H 2 . 3.7 Analyze. Given a box diagram and formulas of reactants, draw a box diagram of products. Plan. Write and balance the chemical equation. Determine combining ratios of elements and decide on limiting reactant. Draw a box diagram of products, containing the correct number of product molecules and only excess reactant. Solve. N2 + 3H2 3NH3. N2 = , NH3 = Each N atom (1/2 of an N 2 molecule) reacts with 3 H atoms (1.5 H 2 molecules) to form an NH 3 molecule. Eight N atoms (4 N 2 molecules) require 24 H atoms (12 H 2 39 3 Stoichiometry Solutions to Exercises molecules) for complete reaction. Only 9 H 2 molecules are available, so H 2 is the limiting reactant. Nine H 2 molecules (18 H atoms) determine that 6 NH 3 molecules are produced. One N 2 molecule is in excess. Check. Verify that mass is conserved in your solution, that the number and kinds of atoms are the same in reactant and product diagrams. In this example, there are 8 N atoms and 18 H atoms in both diagrams, so mass is conserved. 3.8 Each NO molecule reacts with 1 O atom (1/2 of an O 2 molecule) to produce 1 NO 2 molecule. Eight NO molecules react with 8 O atoms (4 O 2 molecules) to produce 8 NO 2 molecules. One O 2 molecule doesn’t react (is in excess). NO is the limiting reactant. (b) % yield = actual yield theoretical yield × 100; actual yield = % yield 100 × theoretical yield The theoretical yield from part (a) is 8 NO 2 molecules. If the percent yield is 75%, then 0.75(8) = 6 NO 2 would appear in the products box. Balancing Chemical Equations 3.9 (a) In balancing chemical equations, the law of conservation of mass, that atoms are neither created nor destroyed during the course of a reaction, is observed. This means that the number and kinds of atoms on both sides of the chemical equation must be the same. Subscripts in chemical formulas should not be changed when balancing equations, because changing the subscript changes the identity of the compound (law of constant composition). liquid water = H2O(l); water vapor = H2O(g); aqueous sodium chloride = NaCl(aq); solid sodium chloride = NaCl(s) (b) (c) 40 3 Stoichiometry 3.10 (a) Solutions to Exercises In a CO molecule, there is one O atom bound to C. 2CO indicates that there are two CO molecules, each of which contains one C and one O atom. Adding a subscript 2 to CO to form CO 2 means that there are two O atoms bound to one C in a CO2 molecule. The composition of the different molecules, CO 2 and CO, is different and the physical and chemical properties of the two compounds they constitute are very different. The subscript 2 changes molecular composition and thus properties of the compound. The prefix 2 indicates how many molecules (or moles) of the original compound are under consideration. Yes. There are the same number and kinds of atoms on the reactants side and the products side of the equation. 2CO(g) + O 2 (g) → 2CO 2 (g) N 2 O 5 (g) + H 2 O(l) → 2HNO 3 (aq) CH 4 (g) + 4Cl 2 (g) → CCl 4 (l) + 4HCl(g) Al 4 C 3 (s) + 12H 2 O(l) → 4Al(OH) 3 (s) + 3CH 4 (g) 2C 5 H 1 0O 2 (l) + 13O 2 (g) → 10CO 2 (g) + 10H 2 O(l) 2Fe(OH) 3 (s) + 3H 2 SO 4 (aq) → Fe 2 (SO 4 ) 3 (aq) + 6H 2 O(l) Mg 3 N 2 (s) + 4H 2 SO 4 (aq) → 3MgSO 4 (aq) + (NH 4 ) 2 SO 4 (aq) 6Li(s) + N 2 (g) → 2Li 3 N(s) La 2 O 3 (s) + 3H 2 O(l) → 2La(OH) 3 (aq) 2NH 4 NO 3 (s) → 2N 2 (g) + O 2 (g) + 4H 2 O(g) Ca 3 P 2 (s) + 6H 2 O(l) → 3Ca(OH) 2 (aq) + 2PH 3 (g) 3Ca(OH) 2 (aq) + 2H 3 PO 4 (aq) → Ca 3 (PO 4 ) 2 (s) + 6H 2 O(l) 2AgNO 3 (aq) + Na 2 SO 4 (aq) → Ag 2 SO 4 (s) + 2NaNO 3 (aq) 4CH 3 NH 2 (g) + 9O 2 (g) → 4CO 2 (g) + 10H 2 O(g) + 2N 2 (g) CaC 2 (s) + 2H 2 O(l) → Ca(OH) 2 (aq) + C 2 H 2 (g) Δ 2KClO 3 (s ) → 2KCl(s) + 3O 2 (g ) (b) 3.11 (a) (b) (c) (d) (e) (f) (g) 3.12 (a) (b) (c) (d) (e) (f) (g) 3.13 (a) (b) (c) (d) (e) Zn(s) + H 2 SO 4 (aq) → H 2 (g) + ZnSO 4 (aq) PCl 3 (l) + 3H 2 O(l) → H 3 PO 3 (aq) + 3HCl(aq) 3H 2 S(g) + 2Fe(OH) 3 (s) → Fe 2 S 3 (s) + 6H 2 O(g) SO 3 (g) + H 2 O(l) → H 2 SO 4 (aq) B 2 S 3 (s) + 6H 2 O(l) → 2H 3 BO 3 (aq) + 3H 2 S(g) Pb(NO3)2(aq) + 2 NaI(aq) → 2 NaNO3 (aq) + PbI2(s) Δ 2Hg(NO3 )2 (s) → 2HgO(s) + 4 NO2 (g) + O 2 (g) 3.14 (a) (b) (c) (d) 41 3 Stoichiometry (e) Solutions to Exercises Cu(s) + 2H 2 SO 4 (aq) → CuSO 4 (aq) + SO 2 (g) + 2H 2 O(l) Patterns of Chemical Reactivity 3.15 (a) When a metal reacts with a nonmetal, an ionic compound forms. The combining ratio of the atoms is such that the total positive charge on the metal cation(s) is equal to the total negative charge on the nonmetal anion(s). Determine the formula by balancing the positive and negative charges in the ionic product. All ionic compounds are solids. 2 Na(s) + Br 2 (l) → 2NaBr(s) The second reactant is oxygen gas from the air, O 2 (g). The products are CO 2 (g) and H 2 O(l). 2C 6 H 6 (l) + 15O 2 (g) → 12CO 2 (g) + 6H 2 O(l). Neutral Ca atom loses 2e – to form Ca 2 +. Neutral O 2 molecule gains 4e – to form 2O 2 –. The formula of the product will be CaO, because the cationic and anionic charges are opposite and equal. 2Ca(s) + O 2 (g) → 2CaO The products are CO 2 (g) and H 2 O(l). C 3 H 6 O(l) + 4O 2 (g) → 3CO 2 (g) + 3H 2 O(l) Mg(s) + Cl 2 (g) → MgCl 2 (s) Δ BaCO 3 (s ) → BaO(s) + CO 2 (g) (b) 3.16 (a) (b) 3.17 (a) (b) (c) (d) 3.18 (a) (b) (c) (d) 3.19 (a) (b) (c) (d) (e) 3.20 (a) (b) (c) (d) (e) C 8 H 8 (l) + 10O 2 (g) → 8CO 2 (g) + 4H 2 O(l) CH 3 OCH 3 is C 2 H 6 O. C 2 H 6 O(l) + 3O 2 (g) → 2CO 2 (g) + 3H 2 O(l) 2Al(s) + 3O2 (g) → Al2O3(s) Δ Cu(OH) 2 (s) → CuO(s) + H 2 O(g) C 7 H 1 6(l) + 11O 2 (g) → 7CO 2 (g) + 8H 2 O(l) 2C 5 H 1 2O(l) + 15O 2 (g) → 10CO 2 (g) + 12H 2 O(l) 2Al(s) + 3Cl 2 (g) → 2AlCl 3 (s) combination combustion C 2 H 4 (g) + 3O 2 (g) → 2CO 2 (g) + 2H 2 O(l) 6Li(s) + N 2 (g) → 2Li 3 N(s) combination PbCO 3 (s) → PbO(s) + CO 2 (g) decomposition C 7 H 8 O 2 (l) + 8O 2 (g) → 7CO 2 (g) + 4H 2 O(l) combustion 2C 3 H 6 (g) + 9O 2 (g) → 6CO 2 (g) + 6H 2 O(l) combustion NH 4 NO 3 (s) → N 2 O(g) + 2H 2 O(l) decomposition C 5 H 6 O(l) + 6O 2 (g) → 5CO 2 (g) + 3H 2 O(l) N 2 (g) + 3H 2 (g) → 2NH 3 (g) combination combustion K 2 O(s) + H 2 O(l) → 2KOH(aq) combination 42 3 Stoichiometry Formula Weights 3.21 Solutions to Exercises Analyze. Given molecular formula or name, calculate formula weight. Plan. If a name is given, write the correct molecular formula. Then, follow the method in Sample Exercise 3.5. Solve. (a) (b) (c) (d) (e) (f) (g) HNO3: 1(1.0) + 1(14.0) + 3(16.0) = 63.0 amu KMnO4: 1(39.1) + 1(54.9) + 4(16.0) = 158.0 amu Ca3(PO4)2: 3(40.1) + 2(31.0) + 8(16.0) = 310.3 amu SiO2: 1(28.1) + 2(16.0) = 60.1 amu Ga2S3: 2(69.7) + 3(32.1) = 235.7 amu Cr2 (SO 4 ) 3 : 2(52.0) + 3(32.1) + 12(16.0) = 392.3 amu PCl3: 1(31.0) + 3(35.5) = 137.5 amu 3.22 Formula weight in amu to 1 decimal place. (a) (b) (c) (d) (e) N 2 O: FW = 2(14.0) + 1(16.0) = 44.0 amu HC 7 H 5 O 2 : 7(12.0) + 6(1.0) + 2(16.0) = 122.0 amu Mg(OH) 2 : 1(24.3) + 2(16.0) + 2(1.0) = 58.3 amu (NH 2 ) 2 CO: 2(14.0) + 4(1.0) + 1(12.0) + 1(16.0) = 60.0 amu CH 3 CO 2 C 5 H 1 1: 7(12.0) + 14(1.0) + 2(16.0) = 130.0 amu 3.23 Plan. Calculate the formula weight (FW), then the mass % oxygen in the compound. Solve. (a) C17H19NO3: FW = 17(12.0) + 19(1.0) + 1(14.0) + 3(16.0) = 285.0 amu %O = 3(16.0) amu × 100 = 16.842 = 16.8% 285.0 amu (b) C18H 2 1 NO 3 : FW = 18(12.0) + 21(1.0) + 1(14.0) + 3(16.0) = 299.0 amu %O = 3(16.0) amu × 100 = 16.054 = 16.1% 299.0 amu (c) C17H21NO 4 : FW = 17(12.0) + 21(1.0) + 1(14.0) + 4(16.0) = 303.0 amu %O = 4(16.0) amu × 100 = 21.122 = 21.1% 303.0 amu (d) C22H24N2O8: FW = 22(12.0) + 24(1.0) + 2(14.0) + 8(16.0) = 444.0 amu %O = 8(16.0) amu × 100 = 28.829 = 28.8% 444.0 amu (e) C41H64O13: FW = 4(12.0) + 64(1.0) + 13(16.0) = 764.0 amu %O = 13(16.0) amu × 100 = 27.225 = 27.2% 764 amu 43 3 Stoichiometry (f) %O = 24(16.0) amu × 100 = 26.519 = 26.5% 1448.0 amu Solutions to Exercises C66H75Cl2N9O24: FW = 66(12.0)+75(1.0)+2(35.5)+9(14.0)+24(16.0) = 1448.0 amu 3.24 (a) C 2 H 2 : FW = 2(12.0) + 2(1.0) = 26.0 amu 2(12.0) amu × 100 = 92.3% 26.0 amu HC 6 H 7 O 6 : FW = 6(12.0) + 8(1.0) + 6(16.0) = 176.0 amu %C = %H = 8(1.0) amu × 100 = 4.5% 176.0 amu (b) (c) (NH 4 ) 2 SO 4 : FW = 2(14.0) + 8(1.0) + 1(32.1) + 4(16.0) = 132.1 amu %H = 8(1.0) amu × 100 = 6.1% 132.1 amu (d) PtCl 2 (NH 3 ) 2 : FW = 1(195.1) + 2(35.5) + 2(14.0) + 6(1.0) = 300.1 amu % Pt = 1(195.1) amu × 100 = 65.01% 300.1 amu (e) C 1 8H 2 4O 2 : FW = 18(12.0) + 24(1.0) + 2(16.0) = 272.0 amu %O = 2(16.0) amu × 100 = 11.8% 272.0 amu (f) C 1 8H 2 7NO 3 : FW = 18(12.0) + 27(1.0) + 1(14.0) + 3(16.0) = 305.0 amu %C = 18(12.0) amu × 100 = 70.8% 305.0 amu 3.25 Plan. Follow the logic for calculating mass % C given in Sample Exercise 3.6. (a) C 7 H 6 O: FW = 7(12.0) + 6(1.0) + 1(16.0) = 106.0 amu %C = 7(12.0) amu × 100 = 79.2% 106.0 amu Solve. (b) C 8 H 8 O 3 : FW = 8(12.0) + 8(1.0) + 3(16.0) = 152.0 amu %C = 8(12.0) amu × 100 = 63.2% 152.0 amu (c) C 7 H 1 4O 2 : FW = 7(12.0) + 14(1.0) + 2(16.0) = 130.0 amu %C = 7(12.0) amu × 100 = 64.6% 130.0 amu 3.26 (a) CO 2 : FW = 1(12.0) + 2(16.0) = 44.0 amu %C = 12.0 amu × 100 = 27.3% 44.0 amu (b) CH 3 OH: FW = 1(12.0) + 4(1.0) + 1(16.0) = 32.0 amu %C = 12.0 amu × 100 = 37.5% 32.0 amu 44 3 Stoichiometry (c) C 2 H 6 : FW = 2(12.0) + 6(1.0) = 30.0 amu %C = 2(12.0) amu × 100 = 80.0% 30.0 amu Solutions to Exercises (d) CS(NH 2 ) 2 : FW = 1(12.0) + 1(32.1) + 2(14.0) + 4(1.0) = 76.1 amu %C = 12.0 amu × 100 = 15.8% 76.1 amu Avogadro’s Number and the Mole 3.27 (a) (b) 3.28 3.29 (a) 6.022 × 10 2 3. This is the number of objects in a mole of anything. The formula weight of a substance in amu has the same numerical value as the molar mass expressed in grams. exactly 12 g (b) 6.0221421 × 10 2 3, Avogadro’s number Plan. Since the mole is a counting unit, use it as a basis of comparison; determine the total moles of atoms in each given quantity. Solve. 23 g Na contains 1 mol of atoms 0.5 mol H 2 O contains (3 atoms × 0.5 mol) = 1.5 mol atoms 6.0 × 10 2 3 N 2 molecules contains (2 atoms × 1 mol) = 2 mol atoms 3.30 3.0 × 10 2 3 H 2 O 2 molecules contains (4 atoms × 0.5 mol) = 2 mol atoms 32 g O 2 contains (2 atoms × 1 mol) = 2 mol atoms 2.0 mol CH 4 contains (5 atoms × 2 mol) = 10 mol atoms 3.31 Analyze. Given: 160 lb/person; Avogadro’s number of people, 6.022 × 10 2 3 people. Find: mass in kg of Avogadro’s number of people; compare with mass of Earth. Plan. people → mass in lb → mass in kg; mass of people /mass of Earth Solve. 6.022 × 10 23 people × 4.370 × 10 25 kg of people 5.98 × 10 24 kg Earth 1 kg 160 lb × = 4.370 × 10 25 = 4.37 × 10 25 or 4.4 × 10 25 kg 2.2046 lb person = 7.31 or 7.3 One mole of people weighs 7.31 times as much as Earth. Check. This mass of people is reasonable since Avogadro’s number is large. Estimate: 160 lb ≈ 70 kg; 6 × 10 2 3 × 70 = 420 × 10 2 3 = 4.2 × 10 2 5 kg 3.32 300 million = 300 × 10 6 = 3.00 × 10 8 or 3 × 10 8 people (The number 300 million has an ambiguous number of sig figs.) 6.022 × 10 23 c / 3.00 × 10 8 people × $1 $6.022 × 10 21 = = 2.007 × 10 13 = $2.01 × 10 13 /person 100 c 3.00 × 10 8 people / $2.007 × 10 13 $1.35 × 10 13 = 1.487 = 1.49 or 1 $13.5 trillion = $1.35 × 10 13 45 3 Stoichiometry 3.33 (a) Solutions to Exercises Each person would receive an amount that is 1.49 (or 1) times the dollar amount of the national debt. Analyze. Given: 0.105 mol sucrose, C12H22O11. Find: mass in g. Plan. Use molar mass (g/mol) of C12H22O11 to find g C12H22O11. Solve. molar mass = 12(12.0107) + 22(1.00794) + 11(15.9994) = 342.296 = 342.30 0.105 mol CaH 2 × 342.30 g 1 mol = 35.942 = 35.9 g C 12 H 22 O 11 Check. 0.1(342) = 34.2 g. The calculated result is reasonable. (b) Analyze. Given: mass. Find: moles. Plan. Use molar mass of Zn(NO 3 ) 2 . Solve. molar mass = 1(65.39) + 2(14.01) + 6(16.00) = 189.41 = 189.4 143.5 g Zn(NO 3 ) 2 × 1 mol = 0.7576 mol Zn(NO 3 ) 2 189.41 g Check. 140/180 ≈ 7/9 = 0.78 mol (c) Analyze. Given: moles. Find: molecules. Plan. Use Avogadro’s number. Solve. 1.0 × 10 −6 mol CH 3 CH 2 OH × 6.022 × 10 23 molecules = 6.022 × 10 17 1 mol = 6.0 × 10 17 CH 3 CH 2 OH molecules Check. (1.0 × 10 − 6 )(6 × 1 0 2 3 ) = 6 × 10 1 7 (d) Analyze. Given: mol NH3. Find: N atoms. Plan. mol NH3 → mol N atoms → N atoms Solve. 0.410 mol NH 3 × 1 mol N atoms 6.022 × 10 23 atoms × 1 mol NH 3 1 mol = 2.47 × 10 23 C atoms Check. (0.4)(6 × 10 2 3) = 2.4 × 10 2 3. 3.34 (a) molar mass = 1(112.41) + 1(32.07) = 144.48 g 5.76 × 10 − 3 mol CdS × 144.48 g 1 mol = 0.832 g CdS (b) molar mass = 1(14.01) + 4(1.008) + 1(35.45) = 53.49 g/mol 112.6 g NH 4 Cl × 1 mol = 2.1051 = 2.11 mol NH 4 Cl 53.49 g (c) 1.305 × 10 − 2 mol C 6 H 6 × 6.02214 × 10 23 molecules = 7.859 × 1021 C6H6 molecules 1 mol 9 mol O 6.022 × 10 23 O atoms × 1 mol Al(NO 3 ) 3 1 mol = 2.64 × 10 22 O atoms (d) 4.88 × 10 − 3 mol Al(NO 3 ) 3 × 46 3 Stoichiometry 3.35 (a) Solutions to Exercises Solve. Analyze/Plan. See Solution 3.33 for stepwise problem-solving approach. (NH 4 ) 3 PO 4 molar mass = 3(14.007) + 12(1.008) + 1(30.974) + 4(16.00) = 149.091 = 149.1 g/mol 2.50 × 10 − 3 mol (NH 4 ) 3 PO 4 × 149.1 g (NH 4 ) 3 PO 4 1 mol = 0.373 g (NH 4 ) 3 PO 4 (b) AlCl3 molar mass = 26.982 + 3(35.453) = 133.341 = 133.34 g/mol 0.2550 g AlCl 3 × 3 mol Cl − 1 mol × = 5.737 × 10 − 3 mol Cl − 133.34 g AlCl 3 1 mol AlCl 3 (c) C 8 H 1 0N 4 O 2 molar mass = 8(12.01) + 10(1.008) + 4(14.01) + 2(16.00) = 194.20 = 194.2 g/mol 7.70 × 10 20 molecules × 194.2 g C 8 H 10 N 4 O 2 1 mol × 23 1 mol caffeine 6.022 × 10 molecules = 0.248 g C8H10N4O2 (d) 3.36 (a) 0.406 g cholesterol 0.00105 mol = 387 g cholesterol/mol Fe 2 (SO 4 ) 3 molar mass = 2(55.845) + 3(32.07) + 12(16.00) = 399.900 = 399.9 g/mol 0.0714 mol Fe 2 (SO 4 )3 × 399.9 g Fe 2 (SO 4 )3 = 28.553 = 28.6 g Fe 2 (SO 4 )3 1 mol (b) (NH 4 ) 2 CO 3 molar mass = 2(14.007) + 8(1.008) + 12.011 + 3(15.9994) = 96.0872 = 96.087 g/mol 8.776 g (NH 4 ) 2 CO 3 × 2 mol NH 4 + 1 mol × = 0.1827 mol NH 4 + 1 mol (NH 4 ) 2 CO 3 96.087 g (NH 4 ) 2 CO 3 (c) C 9 H 8 O 4 molar mass = 9(12.01) + 8(1.008) + 4(16.00) = 180.154 = 180.2 g/mol 6.52 × 10 21 molecules × (d) 3.37 (a) 15.86 g Valium 0.05570 mol 1 mol 6.022 × 10 23 molecules × 180.2 g C 9 H 8 O 4 1 mol aspirin = 1.95 g C 9 H 8 O 4 = 284.7 g Valium/mol C 6 H 1 0OS 2 molar mass = 6(12.01) + 10(1.008) + 1(16.00) + 2(32.07) = 162.28 = 162.3 g/mol (b) Plan. mg → g → mol 5.00 mg allicin × 1 mg Solve. × 1 mol = 3.081 × 10 − 5 = 3.08 × 10 − 5 mol allicin 162.3 g 1 × 10 −3 g Check. 5.00 mg is a small mass, so the small answer is reasonable. (5 × 10 – 3)/200 = 2.5 × 10 – 5 47 3 Stoichiometry (c) Solve. 3.081 × 10 − 5 mol allicin × mol Solutions to Exercises 6.022 × 10 23 molecules Plan. Use mol from part (b) and Avogadro’s number to calculate molecules. = 1.855 × 10 19 = 1.86 × 10 19 allicin molecules Check. (3 × 10 – 5)(6 × 10 2 3) = 18 × 10 1 8 = 1.8 × 10 1 9 (d) Plan. Use molecules from part (c) and molecular formula to calculate S atoms. Solve. 1.855 × 10 19 allicin molecules × 2 S atoms = 3.71 × 10 19 S atoms 1 allicin molecule Check. Obvious. 3.38 (a) C 1 4H 1 8N 2 O 5 molar mass = 14(12.01) + 18(1.008) + 2(14.01) + 5(16.00) = 294.30 g/mol (b) (c) 1.00 mg aspartame × 1 × 10 g 1 mg −3 × 1 mol = 3.398 × 10 −6 = 3.40 × 10 −6 mol aspartame 294.3 g 3.398 × 10 −6 mol aspartame × 6.022 × 10 23 molecules = 2.046 × 10 18 1 mol = 2.05 × 1018 aspartame molecules 18 H atoms = 3.68 × 10 19 H atoms 1 aspartame molecule (d) 3.39 (a) 2.046 × 10 18 aspartame molecules × Analyze. Given: C 6 H 1 2O 6 , 1.250 × 10 2 1 C atoms. Find: H atoms. Plan. Use molecular formula to determine number of H atoms that are present with 1.250 × 10 2 1 C atoms. Solve. 12 H atoms 2 H = × 1.250 × 10 21 C atoms = 2.500 × 10 21 H atoms 6 C atoms 1C Check. (2 × 1 × 10 2 1) = 2 × 10 2 1 (b) Plan. Use molecular formula to find the number of glucose molecules that contain 1.250 × 10 2 1 C atoms. Solve. 1 C 6 H 12 O 6 molecule 6 C atoms × 1.250 × 10 21 C atoms = 2.0833 × 10 20 = 2.083 × 10 20 C 6 H 12 O 6 molecules Check. (12 × 10 2 0/6) = 2 × 10 2 0 (c) Plan. Use Avogadro’s number to change molecules → mol. Solve. 2.0833 × 10 20 C 6 H 12 O 6 molecules × 1 mol 6.022 × 10 23 molecules = 3.4595 × 10 − 4 = 3.460 × 10 − 4 mol C 6 H 12 O 6 Check. (2 × 10 2 0)/(6 × 10 2 3) = 0.33 × 10 – 3 = 3.3 × 10 – 4 (d) Plan. Use molar mass to change mol → g. Solve. 1 mole of C 6 H 1 2O 6 weighs 180.0 g (Sample Exercise 3.9) 48 3 Stoichiometry 3.4595 × 10 − 4 mol C 6 H 12 O 6 × 1 mol Solutions to Exercises 180.0 g C 6 H 12 O 6 = 0.06227 g C 6 H 12 O 6 Check. 3.5 × 180 = 630; 630 × 10 – 4 = 0.063 3.40 (a) (b) 7.08 × 10 20 H atoms × 7.08 × 10 20 H atoms × 19 C atoms = 4.80 × 10 20 C atoms 28 H atoms 28 H atoms = 2.529 × 10 19 1 C19H28O2 molecule = 2.53 × 1019 C19H28O2 molecules (c) 2.529 × 10 19 C 19 H 28 O 2 molecules × 1 mol 6.022 × 10 23 molecules = 4.199 × 10 − 5 = 4.20 × 10−5 mol C19H28O2 (d) C 1 9H 2 8O 2 molar mass = 19(12.01) + 28(1.008) + 2(16.00) = 288.41 = 288.4 g/mol 4.199 × 10 − 5 mol C 19 H 28 O 2 × 3.41 Analyze. Given: g C 2 H 3 Cl/L. 288.4 g C 19 H 28 O 2 1 mol = 0.0121 g C 19 H 28 O 2 Find: mol/L, molecules/L. Plan. The /L is constant throughout the problem, so we can ignore it. Use molar mass for g → mol, Avogadro’s number for mol → molecules. Solve. 2.0 × 10 −6 g C 2 H 3 Cl 1L × 1 mol C 2 H 3 Cl = 3.20 × 10 − 8 = 3.2 × 10 − 8 mol C 2 H 3 Cl/L 62.50 g C 2 H 3 Cl 3.20 × 10 −8 mol C 2 H 3 Cl 6.022 × 10 23 molecules × = 1.9 × 10 16 molecules/ L 1L 1 mol Check. (200 × 10 – 8)/60 = 2.5 × 10 – 8 mol (2.5 × 10 – 8) × (6 × 10 2 3) = 15 × 10 1 5 = 1.5 × 10 1 6 3.42 25 × 10 − 6 g C 21 H 30 O 2 × 1 mol C 21 H 30 O 2 = 7.95 × 10 − 8 = 8.0 × 10 − 8 mol C 21 H 30 O 2 314.5 g C 21 H 30 O 2 6.022 × 10 23 molecules = 4.8 × 10 16 C 21 H 30 O 2 molecules 1 mol 7.95 × 10 −8 mol C 21 H 30 O 2 × Empirical Formulas 3.43 (a) Analyze. Given: moles. Find: empirical formula. Plan. Find the simplest ratio of moles by dividing by the smallest number of moles present. Solve. 0.0130 mol C / 0.0065 = 2 0.039 mol H / 0.0065 = 6 0.0065 mol O / 0.0065 = 1 49 3 Stoichiometry The empirical formula is C 2 H 6 O. Check. The subscripts are simple integers. (b) Solutions to Exercises Analyze. Given: grams. Find: empirical formula. Plan. Calculate the moles of each element present, then the simplest ratio of moles. Solve. 11.66 g Fe × 5.01 g O × 1 mol Fe = 0.2088 mol Fe; 0.2088 / 0.2088 = 1 55.85 g Fe 1 mol O = 0.3131 mol O; 0.3131 / 0.2088 ≈ 1.5 16.00 g O Multiplying by two, the integer ratio is 2 Fe : 3 O; the empirical formula is Fe 2 O 3 . Check. The subscripts are simple integers. (c) Analyze. Given: mass %. Find: empirical formulas. Plan. Assume 100 g sample, calculate moles of each element, find the simplest ratio of moles. Solve. 40.0 g C × 1 mol C = 3.33 mol C; 3.33 / 3.33 = 1 12.01 g C 1 mol H 6.7 g H × = 6.65 mol H; 6.65 / 3.33 ≈ 2 1.008 mol H 1 mol O 53.3 g O × = 3.33 mol O; 3.33 / 3.33 = 1 16.00 mol O The empirical formula is CH 2 O. Check. The subscripts are simple integers. 3.44 (a) Calculate the simplest ratio of moles. 0.104 mol K / 0.052 = 2 0.052 mol C / 0.052 = 1 0.156 mol O / 0.052 = 3 The empirical formula is K 2 CO 3 . (b) Calculate moles of each element present, then the simplest ratio of moles. 1 mol Sn = 0.04448 mol Sn; 0.04448 / 0.04448 = 1 118.7 g Sn 1 mol F 3.37 g F × = 0.1774 mol F; 0.1774 / 0.04448 ≈ 4 19.00 g FSn The integer ratio is 1 Sn : 4 F; the empirical formula is SnF4 . 5.28 g Sn × (c) Assume 100 g sample, calculate moles of each element, find the simplest ratio of moles. 50 3 Stoichiometry 87.5% N = 87.5 g N × Solutions to Exercises 1 mol N = 6.25 mol N; 6.25 / 6.25 = 1 14.01 g 1 mol 12.5% H = 12.5 g H × = 12.4 mol H; 12.4 / 6.25 ≈ 2 1.008 g The empirical formula is NH 2 . 3.45 Analyze/Plan. The procedure in all these cases is to assume 100 g of sample, calculate the number of moles of each element present in that 100 g, then obtain the ratio of moles as smallest whole numbers. Solve. (a) 10.4 g C × 27.8 g S × 1 mol C = 0.866 mol C; 0.866 / 0.866 = 1 12.01 g C 1 mol S = 0.867 mol S; 0.867 / 0.866 ≈ 1 32.07 g S 1 mol Cl = 1.74 mol Cl; 1.74 / 0.866 ≈ 2 35.45 g Cl 61.7 g Cl × The empirical formula is CSCl 2 . (b) 21.7 g C × 9.6 g O × 1 mol C = 1.81 mol C; 1.81 / 0.600 ≈ 3 12.01 g C 1 mol O = 0.600 mol O; 0.600 / 0.600 = 1 16.00 g O 1 mol F = 3.62 mol F; 3.62 / 0.600 ≈ 6 19.00 g F 68.7 g F × The empirical formula is C 3 OF 6 . (c) 32.79 g Na × 13.02 g Al × 54.19 g F × 1 mol Na = 1.426 mol Na; 1.426 / 0.4826 ≈ 3 22.99 g Na 1 mol Al = 0.4826 mol Al; 0.4826 / 0.4826 = 1 26.98 g Al 1 mol F = 2.852 mol F; 2.852 / 0.4826 ≈ 6 19.00 g F The empirical formula is Na 3 AlF 6 . 3.46 See Solution 3.45 for stepwise problem-solving approach. (a) 1 mol K = 1.414 mol K; 1.414/0.4714 ≈ 3 39.10 g K 1 mol P 14.6 g P × = 0.4714 mol P; 0.4714/0.4714 = 1 30.97 g P 55.3 g K × 30.1 g O × 1 mol O = 1.881 mol O; 1.881/0.4714 ≈ 4 16.00 g O The empirical formula is K 3 PO 4 . 51 3 Stoichiometry (b) 24.5 g Na × 14.9 g Si × 60.6 g F × Solutions to Exercises 1 mol Na = 1.066 mol Na; 1.066/0.5304 ≈ 2 22.99 g Na 1 mol Si = 0.5304 mol si; 0.5304/0.5304 = 1 28.09 Si 1 mol F = 3.189 mol F; 3.189/0.5304 ≈ 6 19.00 g F The empirical formula is Na 2 SiF 6 . (c) 62.1 g C × 5.21 g H × 12.1 g N × 20.7 g O × 1 mol C = 5.17 mol C; 5.17 / 0.864 ≈ 6 12.01 g C 1 mol H = 5.17 mol O; 5.17 / 0.864 ≈ 6 1.008 g H 1 mol N = 0.864 mol N; 0.864 / 0.864 = 1 14.01 g N 1 mol O = 1.29 mol O; 1.29 / 0.864 ≈ 1.5 16.00 g O Multiplying by two, the empirical formula is C 1 2H 1 2N 2 O 3 . 3.47 Analyze. Given: empirical formula, molar mass. Find: molecular formula. Plan. Calculate the empirical formula weight (FW); divide FW by molar mass (MM) to calculate the integer that relates the empirical and molecular formulas. Check. If FW/MM is an integer, the result is reasonable. Solve. (a) MM 84 = =6 FW 14 The subscripts in the empirical formula are multiplied by 6. The molecular formula is C 6 H 1 2. FW CH 2 = 12 + 2(1) = 14. MM 51.5 = =1 FW 51.5 The empirical and molecular formulas are NH 2 Cl. FW NH 2 Cl = 14.01 + 2(1.008) + 35.45 = 51.48. FW HCHO 2 = 12.01 + 1.008 + 2(16.00) = 45.0 MM 90.0 = =2 FW 45.0 (b) 3.48 (a) The molecular formula is H 2 C 2 O 4 . (b) FW C 2 H 4 O = 2(12) + 4(1) + 16 = 44. MM 88 = =2 FW 44 The molecular formula is C 4 H 8 O 2 . 3.49 Analyze. Given: mass %, molar mass. Find: molecular formula. Plan. Use the plan detailed in Solution 3.45 to find an empirical formula from mass % data. Then use the plan detailed in 3.47 to find the molecular formula. Note that some indication of molar mass must be given, or the molecular formula cannot be determined. Check. If there is an integer ratio of moles and MM/ FW is an integer, the result is reasonable. Solve. 52 3 Stoichiometry (a) 92.3 g C × 7.7 g H × Solutions to Exercises 1 mol C = 7.685 mol C; 7.685/7.639 = 1.006 ≈ 1 12.01 g C 1 mol H = 7.639 mol H; 7.639/7.639 = 1 1.008 g H The empirical formula is CH, FW = 13. MM 104 = = 8; the molecular formula is C 8 H 8 . FW 13 (b) 49.5 g C × 5.15 g H × 28.9 g N × 16.5 g O × 1 mol C = 4.12 mol C; 4.12/1.03 ≈ 4 12.01 g C 1 mol H = 5.11 mol H; 5.11/1.03 ≈ 5 1.008 g H 1 mol N = 2.06 mol N; 2.06/1.03 ≈ 2 14.01 g N 1 mol O = 1.03 mol O; 1.03/1.03 = 1 16.00 g O Thus, C 4 H 5 N 2 O, FW = 97. If the molar mass is about 195, a factor of 2 gives the molecular formula C 8 H 1 0N 4 O 2 . (c) 35.51 g C × 4.77 g H × 1 mol C = 2.96 mol C; 2.96/0.592 = 5 12.01 g C 1 mol H = 4.73 mol H; 4.73/0.592 = 7.99 ≈ 8 1.008 g H 1 mol O = 2.37 mol O; 2.37/0.592 = 4 16.00 g O 37.85 g O × 8.29 g N × 1 mol N = 0.592 mol N; 0.592/0.592 = 1 14.01 g N 1 mol Na = 0.592 mol Na; 0.592/0.592 = 1 22.99 g Na 13.60 g Na × The empirical formula is C 5 H 8 O 4 NNa, FW = 169 g. Since the empirical formula weight and molar mass are approximately equal, the empirical and molecular formulas are both NaC 5 H 8 O 4 N. 3.50 Assume 100 g in the following problems. (a) 75.69 g C × 8.80 g H × 1 mol C = 6.30 mol C; 6.30/0.969 = 6.5 12.01 g C 1 mol H = 8.73 mol H; 8.73/0.969 = 9.0 1.008 g H 1 mol O = 0.969 mol O; 0.969/0.969 = 1 16.00 g O 15.51 g O × 53 3 Stoichiometry Solutions to Exercises Multiply by 2 to obtain the integer ratio 13:18:2. The empirical formula is C 1 3H 1 8O 2 , FW = 206 g. Since the empirical formula weight and the molar mass are equal (206 g), the empirical and molecular formulas are C 1 3H 1 8O 2 . (b) 58.55 g C × 13.81 g H × 27.40 g N × 1 mol C = 4.875 mol C; 4.875/1.956 ≈ 2.5 12.01 g C 1 mol H = 13.700 mol H; 13.700/1.956 ≈ 7.0 1.008 g H 1 mol N = 1.956 mol N; 1.956/1.956 = 1.0 14.01 g N Multiply by 2 to obtain the integer ratio 5:14:2. The empirical formula is C 5 H 1 4N 2 ; FW = 102. Since the empirical formula weight and the molar mass are equal (c) (102 g), the empirical and molecular formulas are C 5 H 1 4N 2 . 1 mol C 59.0 g C × = 4.91 mol C; 4.91 / 0.550 ≈ 9 12.01 g C 7.1 g H × 1 mol H = 7.04 mol H; 7.04 / 0.550 ≈ 13 1.008 g H 1 mol O = 1.64 mol O; 1.64 / 0.550 ≈ 3 16.00 g O 26.2 g O × 7.7 g N × 1 mol N = 0.550 mol N; 0.550 / 0.550 = 1 14.01 g N The empirical formula is C 9 H 1 3O 3 N, FW = 183 amu (or g). Since the molecular weight is approximately 180 amu, the empirical formula and molecular formula are the same, C 9 H 1 3O 3 N. 3.51 (a) Analyze. Given: mg CO 2 , mg H 2 O Find: empirical formula of hydrocarbon, CxHy Plan. Upon combustion, all C → CO 2 , all H → H 2 O. mg CO 2 → g CO 2 → mol C; mg H 2 O → g H 2 O, mol H Find simplest ratio of moles and empirical formula. 5.86 × 10 − 3 g CO 2 × 1.37 × 10 − 3 Solve. 1 mol CO 2 1 mol C × = 1.33 × 10 − 4 mol C 44.01 g CO 2 1 mol CO 2 1 mol H 2 O 2 mol H g H2O× × = 1.52 × 10 − 4 mol H 18.02 g H 2 O 1 mol H 2 O Dividing both values by 1.33 × 10 – 4 gives C:H of 1:1.14. This is not “close enough” to be considered 1:1. No obvious multipliers (2, 3, 4) produce an integer ratio. Testing other multipliers (trial and error!), the correct factor seems to be 7. The empirical formula is C 7 H 8 . Check. See discussion of C:H ratio above. 54 3 Stoichiometry (b) Solutions to Exercises Analyze. Given: g of menthol, g CO 2 , g H 2 O, molar mass. Find: molecular formula. Plan/Solve. Calculate mol C and mol H in the sample. 0.2829 g CO 2 × 0.1159 g H 2 O × 1 mol CO 2 1 mol C × = 0.0064281 = 0.006428 mol C 44.01 g CO 2 1 mol CO 2 1 mol H 2 O 2 mol H × = 0.012863 = 0.01286 mol H 18.02 g H 2 O 1 mol H 2 O Calculate g C, g H and get g O by subtraction. 0.0064281 mol C × 0.012863 mol H × 12.01 g C 1 mol C 1 mol H = 0.07720 g C = 0.01297 g H 1.008 g H mass O = 0.1005 g sample – (0.07720 g C + 0.01297 g H) = 0.01033 g O Calculate mol O and find integer ratio of mol C: mol H: mol O. 0.01033 g O × 1 mol O = 6.456 × 10 − 4 mol O 16.00 g O 6.456 × 10 −4 6.456 × 10 − 4 Divide moles by 6.456 × 10 – 4. C: 0.006428 6.456 × 10 − 4 ≈ 10; H : 0.01286 6.456 × 10 − 4 ≈ 20; O : =1 The empirical formula is C 1 0H 2 0O. FW = 10(12) + 20(1) + 16 = 156; M 156 = =1 FW 156 The molecular formula is the same as the empirical formula, C 1 0H 2 0O. Check. The mass of O wasn’t negative or greater than the sample mass; empirical and molecular formulas are reasonable. 3.52 (a) Plan. Calculate mol C and mol H, then g C and g H; get g O by subtraction. Solve. 6.32 × 10 − 3 g CO 2 × 2.58 × 10 − 3 g H 2 O × 1.436 × 10 − 4 2.863 × 10 − 4 1 mol CO 2 1 mol C × = 1.436 × 10 − 4 = 1.44 × 10 − 4 mol C 44.01 g CO 2 1 mol CO 2 1 mol H 2 O × 2 mol H = 2.863 × 10 − 4 = 2.86 × 10 − 4 mol H 18.02 g H 2 O 1 mol H 2 O 12.01 g C mol C × = 1.725 × 10 − 3 g C = 1.73 mg C 1 mol C 1.008 g H mol H × = 2.886 × 10 − 4 g H = 0.289 mg H 1 mol H mass of O = 2.78 mg sample – (1.725 mg C + 0.289 mg H) = 0.77 mg O 0.77 × 10 − 3 g O × 1 mol O = 4.81 × 10 − 5 mol O. Divide moles by 4.81 × 10 − 5 . 16.00 g O 55 3 Stoichiometry C: 1.44 × 10 −4 ≈ 3; H : 4.81 × 10 − 5 Solutions to Exercises 2.86 × 10 −4 4.81 × 10 −5 ≈ 6; O : =1 4.81 × 10 − 5 4.81 × 10 − 5 The empirical formula is C 3 H 6 O. (b) Plan. Calculate mol C and mol H, then g C and g H. In this case, get N by subtraction. Solve. 14.242 × 10 − 3 g CO 2 × 1 mol CO 2 1 mol C × = 3.2361 × 10 − 4 mol C 44.01 g CO 2 1 mol CO 2 1 mol H 2 O 2 mol H × = 4.5136 × 10 − 4 = 4.532 × 10 − 4 mol H 4.083 × 10 − 3 g H 2 O × 18.02 g H 2 O 1 mol H 2 O 12.01 g C = 3.8866 × 10 − 3 g C = 3.8866 mg C 3.2361 × 10 − 4 g mol C × 1 mol H 4.532 × 10 − 4 mol H × 1.008 g H 1 mol H = 0.45683 × 10 − 3 g H = 0.4568 mg H mass of N = 5.250 mg sample – (3.8866 mg C + 0.4568 mg H) = 0.9066 = 0.907 mg N 0.9066 × 10 − 3 g N × 3.24 × 10 −4 6.47 × 10 − 5 1 mol N = 6.47 × 10 − 5 mol N. Divide moles by 6.47 × 10 – 5. 14.01 g N 4.53 × 10 −4 6.47 × 10 − 5 ≈ 7; N : 6.47 × 10 −5 6.47 × 10 − 5 =1 C: ≈ 5; H : The empirical formula is C 5 H 7 N, FW = 81. A molar mass of 160 ± 5 indicates a factor of 2 and a molecular formula of C 1 0H 1 4N 2 . 3.53 Analyze. Given 2.558 g Na 2 CO 3 • xH 2 O, 0.948 g Na 2 CO 3 . Find: x. Plan. The reaction involved is Na 2 CO 3 • xH 2 O(s) → Na 2 CO 3 (s) + xH 2 O(g). Calculate the mass of H 2 O lost and then the mole ratio of Na 2 CO 3 and H 2 O. Solve. g H 2 O lost = 2.558 g sample – 0.948 g Na 2 CO 3 = 1.610 g H 2 O 0.948 g Na 2 CO 3 × 1.610 g H 2 O × 1 mol Na 2 CO 3 = 0.00894 mol Na 2 CO 3 106.0 g Na 2 CO 3 1 mol H 2 O = 0.08935 mol H 2 O 18.02 g H 2 O The formula is Na 2 CO 3 • 10 H 2 O. Check. x is an integer. 3.54 The reaction involved is MgSO 4 • xH 2 O(s) → MgSO 4 (s) + xH 2 O(g). First, calculate the number of moles of product MgSO 4 ; this is the same as the number of moles of starting hydrate. 2.472 g MgSO 4 × 1 mol MgSO 4 120.4 g MgSO 4 × 1 mol MgSO 4 • x H 2 O 1 mol MgSO 4 = 0.02053 mol MgSO 4 • x H 2 O 56 3 Stoichiometry Thus, 5.061 g MgSO 4 • x H 2 O 0.02053 Solutions to Exercises = 246.5 g/mol = FW of MgSO 4 • x H 2 O FW of MgSO 4 • xH 2 O = FW of MgSO 4 + x(FW of H 2 O). 246.5 = 120.4 + x(18.02). x = 6.998. The hydrate formula is MgSO 4 • 7H 2 O. Alternatively, we could calculate the number of moles of water represented by weight loss: (5.061 – 2.472) = 2.589 g H 2 O lost. 2.589 g H 2 O × 1 mol H 2 O mol H 2 O 0.1437 = 0.1437 mol H 2 O; = = 7.000 18.02 g H 2 O mol MgSO 4 0.02053 Again the correct formula is MgSO 4 • 7H 2 O. Calculations Based on Chemical Equations 3.55 The mole ratios implicit in the coefficients of a balanced chemical equation express the fundamental relationship between amounts of reactants and products. If the equation is not balanced, the mole ratios will be incorrect and lead to erroneous calculated amounts of products. The integer coefficients immediately preceding each molecular formula in a chemical equation give information about relative numbers of moles of reactants and products involved in a reaction. Na 2 SiO 3 (s) + 8HF(aq) → H 2 SiF 6 (aq) + 2NaF(aq) + 3H 2 O(l) (a) Analyze. Given: mol Na 2 SiO 3 . Find: mol HF. Plan. Use the mole ratio 8HF:1Na 2 SiO 3 from the balanced equation to relate moles of the two reactants. Solve. 0.300 mol Na 2 SiO 3 × 8 mol HF = 2.40 mol HF 1 mol Na 2 SiO 3 3.56 3.57 Check. Mol HF should be greater than mol Na 2 SiO 3 . (b) Analyze. Given: mol HF. Find: g NaF. Plan. Use the mole ratio 2NaF:8HF to change mol HF to mol NaF, then molar mass to get NaF. Solve. 0.500 mol HF × 2 mol NaF 41.99 g NaF × = 5.25 g NaF 8 mol HF 1 mol NaF Check. (0.5/4) = 0.125; 0.13 × 42 > 4 g NaF (c) Analyze. Given: g HF Find: g Na 2 SiO 3 . ⎛ mol ⎞ Plan. g HF → mol HF ⎜ ⎜ ratio ⎟ → mol Na 2 SiO 3 → g Na 2 SiO 3 ⎟ ⎝ ⎠ The mole ratio is at the heart of every stoichiometry problem. Molar mass is used to change to and from grams. Solve. 0.800 g HF × 1 mol HF 1 mol Na 2 SiO 3 122.1 g Na 2 SiO 3 × × = 0.610 g Na 2 SiO 3 20.01g HF 8 mol HF 1 mol Na 2 SiO 3 Check. 0.8 (120/160) < 0.75 mol 57 3 Stoichiometry 3.58 C 6 H 1 2O 6 (aq) → 2C 2 H 5 OH(aq) + 2CO 2 (g) (a) (b) 0.400 mol C 6 H 12 O 6 × 7.50 g C 2 H 5 OH × Solutions to Exercises 2 mol CO 2 = 0.800 mol CO 2 1 mol C 6 H 12 O 6 1 mol C 2 H 5 OH 1 mol C 6 H 12 O 6 180.2 g C 6 H 12 O 6 × × 46.07 g C 2 H 5 OH 2 mol C 2 H 5 OH 1 mol C 6 H 12 O 6 = 14.7 g C6H12O6 (c) 3.59 (a) (b) 7.50 g C 2 H 5 OH × 44.01 g CO 2 1 mol C 2 H 5 OH 2 mol CO 2 × × = 7.16 g CO 2 46.07 g C 2 H 5 OH 2 mol C 2 H 5 OH 1 mol CO 2 Al(OH) 3 (s) + 3HCl(aq) → AlCl 3 (aq) + 3H 2 O(l) Analyze. Given mass of one reactant, find stoichiometric mass of other reactant and products. Plan. Follow the logic in Sample Exercise 3.16. Calculate mol Al(OH) 3 in 0.500 g Al(OH 3 ) 3 separately, since it will be used several times. Solve. 0.500 g Al(OH) 3 × 6.410 × 10 − 3 1 mol Al(OH) 3 = 6.410 × 10 − 3 = 6.41 × 10 − 3 mol Al(OH) 3 78.00 g Al(OH) 3 36.46 g HCl 3 mol HCl mol Al(OH) 3 × × = 0.7012 = 0.701 g HCl 1 mol Al(OH) 3 1 mol HCl 1 mol HCl 133.34 g AlCl 3 × = 0.8547 1 mol Al(OH) 3 1 mol AlCl 3 = 0.855 g AlCl 3 6.410 × 10 − 3 mol Al(OH) 3 × 18.02 g H 2 O 3 mol H 2 O × = 0.3465 = 0.347 g H 2 O 1 mol Al(OH) 3 1 mol H 2 O (c) 6.410 × 10 − 3 molAl(OH) 3 × (d) Conservation of mass: mass of products = mass of reactants reactants: Al(OH) 3 + HCl, 0.500 g + 0.701 g = 1.201 g products: AlCl 3 + H 2 O, 0.855 g + 0.347 g = 1.202 g The 0.001 g difference is due to rounding (0.8547 + 0.3465 = 1.2012). This is an excellent check of results. 3.60 (a) (b) Fe 2 O 3 (s) + 3CO(g) → 2Fe(s) + 3CO 2 (g) 0.150 kg Fe 2 O 3 × 1000 g 1 kg × 1 mol Fe 2 O 3 = 0.9393 = 0.939 mol Fe 2 O 3 159.688 g Fe 2 O 3 0.9393 mol Fe 2 O 3 × 28.01 g CO 3 mol CO × = 78.929 = 78.9 g CO 1 mol Fe 2 O 3 1 mol CO 55.845 g Fe 2 mol Fe × = 104.914 = 105 g Fe 1 mol Fe 2 O 3 1 mol Fe 3 mol CO 2 44.01 g CO 2 × = 124.015 = 124 g CO 2 1 mol Fe 2 O 3 1 mol CO 2 (c) 0.9393 mol Fe 2 O 3 × 0.9393 mol Fe 2 O 3 × 58 3 Stoichiometry (d) Mass is conserved. 3.61 (a) (b) Al 2 S 3 (s) + 6H 2 O(l) → 2Al(OH) 3 (s) + 3H 2 S(g) Solutions to Exercises reactants: 150 g Fe 2 O 3 + 78.9 g CO = 228.9 = 229 g products: 104.9 g Fe + 124.0 g CO 2 = 228.9 = 229 g Plan. g A → mol A → mol B → g B. See Solution 3.57 (c). 14.2 g Al 2 S 3 × Solve. 1 mol Al 2 S 3 2 mol Al(OH) 3 78.00 g Al(OH) 3 × × = 14.7 g Al(OH) 3 150.2 g Al 2 S 3 1 mol Al 2 S 3 1 mol Al(OH) 3 ⎛ 2 × 78 ⎞ Check. 14 ⎜ ⎟ ≈ 14(1) ≈ 14 g Al(OH) 3 ⎝ 150 ⎠ 3.62 (a) (b) CaH 2 (s) + 2H 2 O(l) → Ca(OH) 2 (aq) + 2H 2 (g) 8.500 g H 2 × 1 mol H 2 1 mol CaH 2 42.10 g CaH 2 × × = 88.75 g CaH 2 2.016 g H 2 2 mol H 2 1 mol CaH 2 3.63 (a) Analyze. Given: mol NaN 3 . Find: mol N 2 . Plan. Use mole ratio from balanced equation. 1.50 mol NaN 3 × 3 mol N 2 = 2.25 mol N 2 2 mol NaN 3 Solve. Check. The resulting mol N 2 should be greater than mol NaN 3 , (the N 2 :NaN 3 ratio is > 1), and it is. (b) Analyze. Given: g N 2 Find: g NaN 3 . Plan. Use molar masses to get from and to grams, mol ratio to relate moles of the two substances. Solve. 10.0 g N 2 × 1 mol N 2 2 mol NaN 3 65.01 g NaN 3 × × = 15.5 g NaN 3 28.01 g N 2 3 mol N 2 1 mol NaN 3 Check. Mass relations are less intuitive than mole relations. Estimating the ratio of molar masses is sometimes useful. In this case, 65 g NaN 3 /28 g N 2 ≈ 2.25 Then, (10 × 2/3 × 2.25) ≈ 14 g NaN 3 . The calculated result looks reasonable. (c) Analyze. Given: vol N 2 in ft 3 , density N 2 in g/L. Find: g NaN 3 . Plan. First determine how many g N 2 are in 10.0 ft 3 , using the density of N 2 . Then proceed as in part (b). Solve. ( 2.54) 3 cm 3 (12) 3 in 3 1.25 g 1L × × × × 10.0 ft 3 = 354.0 = 354 g N 2 1L 1000 cm 3 1 in 3 1 ft 3 1 mol N 2 2 mol NaN 3 65.01 g NaN 3 354.0 g N 2 × × × = 548 g NaN 3 28.01 g N 2 3 mol N 2 1 mol NaN 3 Check. 1 ft 3 ~ 28 L; 10 ft 3 ~ 280 L; 280 L × 1.25 ~ 350 g N 2 Using the ratio of molar masses from part (b), (350 × 2/3 × 2.25) ≈ 525 g NaN 3 59 3 Stoichiometry 3.64 2C 8 H 1 8(l) + 25O 2 (g) → 16CO 2 (g) + 18H 2 O(l) (a) (b) (c) 1.25 mol C 8 H 18 × 10.0 g C 8 H 18 × Solutions to Exercises 25 mol O 2 = 15.625 = 15.6 mol O 2 2 mol C 8 H 18 32.00 g O 2 1 mol C 8 H 18 25 mol O 2 × × = 35.0 g O 2 114.2 g C 8 H 18 2 mol C 8 H 18 1 mol O 2 3.7854 L 1000 mL 0.692 g × × = 2619.5 = 2.62 × 10 3 g C 8 H 18 1 gal 1L 1 mL 32.00 g O 2 1 mol C 8 H 18 25 mol O 2 × × = 9 ,175.1 g 114.2 g C 8 H 18 2 mol C 8 H 18 1 mol O 2 = 9.18 × 10 3 g O 2 1.00 gal C 8 H 18 × 2.6195 × 10 3 g C 8 H 18 × 3.65 (a) Analyze. Given: dimensions of Al foil. Find: mol Al. density Plan. Dimens ions → vol ⎯⎯ ⎯ ⎯ → mass ⎯molar → mol Al ⎯ ⎯⎯ ⎯ mass Solve. 1.00 cm × 1.00 cm × 0.550 mm × 0.0550 cm 3 Al × 2.699 g Al 1 cm 3 1 cm = 0.0550 cm 3 Al 10 mm × 1 mol Al = 5.502 × 10 − 3 = 5.50 × 10 − 3 mol Al 26.98 g Al Check. 2.699/26.98 ≈ 0.1; (0.055 cm 3 × 0.1) = 5.5 × 10 – 3 mol Al (b) Plan. Write the balanced equation to get a mole ratio; change mol Al → mol AlBr 3 → g AlBr 3 . Solve. 2Al(s) + 3Br 2 (l) → 2AlBr 3 (s) 5.502 × 10 − 3 mol Al × 2 mol AlBr3 266.69 g AlBr3 × = 1.467 = 1.47 g AlBr3 2 mol Al 1 mol AlBr3 Check. (0.006 × 1 × 270) ≈ 1.6 g AlBr 3 3.66 (a) Plan. Calculate a “mole ratio” between nitroglycerine and total moles of gas produced. (12 + 6 + 1 + 10) = 29 mol gas; 4 mol nitro: 29 total mol gas. Solve. 1.592 g 1 mol nitro 29 mol gas × × = 0.10165 = 0.102 mol gas mL 227.1 g nitro 4 mol nitro 55 L 0.10165 mol gas × = 5.5906 = 5.6 L mol 2.00 mL nitro × 2.00 mL nitro × 1.592 g 28.01 g N 2 6 mol N 2 1 mol nitro × × × = 0.589 g N 2 mL 227.1 g nitro 4 mol nitro 1 mol N 2 (b) (c) Limiting Reactants; Theoretical Yields 3.67 (a) The limiting reactant determines the maximum number of product moles resulting from a chemical reaction; any other reactant is an excess reactant. 60 3 Stoichiometry (b) (c) Solutions to Exercises The limiting reactant regulates the amount of products, because it is completely used up during the reaction; no more product can be made when one of the reactants is unavailable. Combining ratios are molecule and mole ratios. Since different molecules have different masses, equal masses of different reactants will not have equal numbers of molecules. By comparing initial moles, we compare numbers of available reactant molecules, the fundamental combining units in a chemical reaction. Theoretical yield is the maximum amount of product possible, as predicted by stoichiometry, assuming that the limiting reactant is converted entirely to product. Actual yield is the amount of product actually obtained, less than or equal to the theoretical yield. Percent yield is the ratio of (actual yield to theoretical yield) × 100. 3.68 (a) (b) No reaction is perfect. Not all reactant molecules come together effectively to form products; alternative reaction pathways may produce secondary products and reduce the amount of desired product actually obtained, or it might not be possible to completely isolate the desired product from the reaction mixture. In any case, these factors reduce the actual yield of a reaction. No, 110% actual yield is not possible. Theoretical yield is the maximum possible amount of pure product, assuming all available limiting reactant is converted to product, and that all product is isolated. If an actual yield of 110% if obtained, the product must contain impurities which increase the experimental mass. Each bicycle needs 2 wheels, 1 frame, and 1 set of handlebars. A total of 4815 wheels corresponds to 2407.5 pairs of wheels. This is more than the number of frames or handlebars. The 2255 handlebars determine that 2255 bicycles can be produced. 2305 frames – 2255 bicycles = 50 frames left over 2407.5 pairs of wheels – 2255 bicycles = 152.5 pairs of wheels left over 2(152.5) = 305 wheels left over (c) 3.69 (a) (b) (c) The handlebars are the “limiting reactant” in that they determine the number of bicycles that can be produced. 40 ,875 L beverage × 1 bottle = 115 ,140.85 = 1.15 × 10 5 portions of beverage 0.355 L 3.70 (a) (The uncertainty in 355 mL limits the precision of the number of portions we can reasonably expect to deliver to three significant figures.) 121,515 bottles; 122,500 caps; 1.15 × 10 5 bottles can be filled and capped. (b) 122,500 caps – 115,141 portions = 7,359 = 7 × 10 3 caps remain 121,515 empty bottles – 115,141 portions = 6374 = 6 × 10 3 bottles remain (Uncertainty in the number of portions delivered limits the results to 1 sig fig.) 61 3 Stoichiometry (c) 3.71 The volume of beverage limits production. Solutions to Exercises Analyze. Given: 1.85 mol NaOH, 1.00 mol CO 2 . Find: mol Na 2 CO 3 . Plan. Amounts of more than one reactant are given, so we must determine which reactant regulates (limits) product. Then apply the appropriate mole ratio from the balanced equation. Solve. The mole ratio is 2NaOH:1CO 2 , so 1.00 mol CO 2 requires 2.00 mol NaOH for complete reaction. Less than 2.00 mol NaOH are present, so NaOH is the limiting reactant. 1.85 mol NaOH × 1 mol Na 2 CO 3 = 0.925 mol Na 2 CO 3 can be produced 2 mol NaOH The Na 2 CO 3 :CO 2 ratio is 1:1, so 0.925 mol Na 2 CO 3 produced requires 0.925 mol CO 2 consumed. (Alternately, 1.85 mol NaOH × 1 mol CO 2 /2 mol NaOH = 0.925 mol CO 2 reacted). 1.00 mol CO 2 initial – 0.925 mol CO 2 reacted = 0.075 mol CO 2 remain. Check. initial change (reaction) final 2NaOH(s) 1.85 mol –1.85 mol 0 mol + CO 2 (g) 1.00 mol –0.925 mol 0.075 mol → Na 2 CO 3 (s) 0 mol +0.925 mol 0.925 mol + H 2 O(l) Note that the “change” line (but not necessarily the “final” line) reflects the mole ratios from the balanced equation. 3.72 0.500 mol Al(OH) 3 × 3 mol H 2 SO 4 = 0.750 mol H 2 SO 4 needed for complete reaction 2 mol Al(OH) 3 Only 0.500 mol H 2 SO 4 available, so H 2 SO 4 limits. 0.500 mol H 2 SO 4 × 0.500 mol H 2 SO 4 × 1 mol Al 2 (SO 4 ) 3 = 0.1667 = 0.167 mol Al 2 (SO 4 ) 3 can form 3 mol H 2 SO 4 2 mol Al(OH) 3 = 0.3333 = 0.333 mol Al(OH) 3 react 3 mol H 2 SO 4 0.500 mol Al(OH) 3 initial – 0.333 mol react = 0.167 mol Al(OH) 3 remain 3.73 3NaHCO 3 (aq) + H 3 C 6 H 5 O 7 (aq) → 3CO 2 (g) + 3H 2 O(l) + Na 3 C 6 H 5 O 7 (aq) (a) Analyze/Plan. Abbreviate citric acid as H 3 Cit. Follow the approach in Sample Exercise 3.19. Solve. 1.00 g NaHCO 3 × 1 mol NaHCO 3 = 1.190 × 10 − 2 = 1.19 × 10 − 2 mol NaHCO 3 84.01 g NaHCO 3 1 mol H 3 Cit = 5.206 × 10 − 3 = 5.21 × 10 − 3 mol H 3 Cit 192.1 g H 3 Cit 1.00 g H 3 C 6 H 5 O 7 × But NaHCO 3 and H 3 Cit react in a 3:1 ratio, so 5.21 × 10 – 3 mol H 3 Cit require 3(5.21 × 10 – 3) = 1.56 × 10 – 2 mol NaHCO 3 . We have only 1.19 × 10 – 2 mol NaHCO 3 , so NaHCO 3 is the limiting reactant. 62 3 Stoichiometry (b) (c) 1.190 × 10 − 2 mol NaHCO 3 × 1.190 × 10 − 2 mol NaHCO 3 × Solutions to Exercises 44.01 g CO 2 3 mol CO 2 × = 0.524 g CO 2 3 mol NaHCO 3 1 mol CO 2 1 mol H 3 Cit = 3.968 × 10 − 3 3 mol NaHCO 3 = 3.97 × 10 − 3 mol H 3 Cit react 5.206 × 10 – 3 mol H 3 Cit – 3.968 × 10 – 3 mol react = 1.238 × 10 – 3 = 1.24 × 10 – 3 mol H 3 Cit remain 1.238 × 10 − 3 mol H 3 Cit × 192.1 g H 3 Cit = 0.238 g H 3 Cit remain mol H 3 Cit 3.74 4NH 3 (g) + 5O 2 (g) → 4NO(g) + 6H 2 O(g) (a) Follow the approach in Sample Exercise 3.19. 1.50 g NH 3 × 2.75 g O 2 × 1 mol NH 3 = 0.08808 = 0.0881 mol NH 3 17.03 g NH 3 1 mol O 2 = 0.08594 = 0.0859 mol O 2 32.00 g O 2 4 mol NH 3 = 0.06875 = 0.0688 mol NH 3 required 5 mol O 2 0.08594 mol O 2 × More than 0.0688 mol NH 3 is available, so O 2 is the limiting reactant. (b) 0.08594 mol O 2 × 4 mol NO 30.01 g NO × = 2.063 = 2.06 g NO produced 5 mol O 2 1 mol NO 0.08594 mol O 2 × 6 mol H 2 O 18.02 g H 2 O × = 1.8583 = 1.86 g H 2 O produced 1 mol H 2 O 5 mol O 2 (c) 0.08808 mol NH 3 – 0.06875 mol NH 3 reacted = 0.01933 = 0.0193 mol NH 3 remain 0.01933 mol NH 3 × 17.03 g NH 3 1 mol NH 3 = 0.32919 = 0.329 g NH 3 remain (d) mass products = 2.06 g NO + 1.86 g H 2 O + 0.329 g NH 3 remaining = 4.25 g products mass reactants = 1.50 g NH 3 + 2.75 g O 2 = 4.25 g reactants (For comparison purposes, the mass of excess reactant can be either added to the products, as above, or subtracted from reactants.) 3.75 Analyze. Given: initial g Na 2 CO 3 , g AgNO 3 . Find: final g Na 2 CO 3 , AgNO 3 , Ag 2 CO 3 , NaNO 3 Plan. Write balanced equation; determine limiting reactant; calculate amounts of excess reactant remaining and products, based on limiting reactant. Solve. 2AgNO 3 (aq) + Na 2 CO 3 (aq) → Ag 2 CO 3 (s) + 2NaNO 3 (aq) 63 3 Stoichiometry 3.50 g Na 2CO3 × 5.00 g AgNO 3 × 1 mol AgNO 3 169.9 g AgNO 3 Solutions to Exercises 1 mol Na 2CO3 = 0.03302 = 0.0330 mol Na 2CO3 106.0 g Na 2CO3 = 0.02943 = 0.0294 mol AgNO 3 0.02943 mol AgNO 3 × 1 mol Na 2 CO 3 = 0.01471 = 0.0147 mol Na 2 CO 3 required 2 mol AgNO 3 AgNO 3 is the limiting reactant and Na 2 CO 3 is present in excess. 2AgNO 3 (aq) initial reaction final 0.0294 mol –0.0294 mol 0 mol + Na 2 CO 3 (aq) 0.0330 mol –0.0147 mol 0.0183 mol → Ag 2 CO 3 (s) 0 mol +0.0147 mol 0.0147 mol + 2NaNO 3 (aq) 0 mol +0.0294 mol 0.0294 mol 0.01830 mol Na 2 CO 3 × 106.0 g/mol = 1.940 = 1.94 g Na 2 CO 3 0.01471 mol Ag 2 CO 3 × 275.8 g/mol = 4.057 = 4.06 g Ag 2 CO 3 0.02943 mol NaNO 3 × 85.00 g/mol = 2.502 = 2.50 g NaNO 3 Check. The initial mass of reactants was 8.50 g, and the final mass of excess reactant and products is 13.50 g; mass is conserved. 3.76 Plan. Write balanced equation; determine limiting reactant; calculate amounts of excess reactant remaining and products, based on limiting reactant. Solve. H 2 SO 4 (aq) + Pb(C 2 H 3 O 2 ) 2 (aq) → PbSO 4 (s) + 2HC 2 H 3 O 2 (aq) 7.50 g H 2 SO 4 × 1 mol H 2 SO 4 = 0.07646 = 0.0765 mol H 2 SO 4 98.09 g H 2 SO 4 1 mol Pb(C 2 H 3 O 2 ) 2 = 0.023056 = 0.0231 mol Pb(C 2 H 3 O 2 ) 2 325.3 g Pb(C 2 H 3 O 2 ) 2 7.50 g Pb(C 2 H 3 O 2 ) 2 × 1 mol H 2 SO 4 :1 mol Pb(C 2 H 3 O 2 ) 2 , so Pb(C 2 H 3 O 2 ) 2 is the limiting reactant. 0 mol Pb(C 2 H 3 O 2 ) 2 , (0.07646 – 0.023056) = 0.0534 mol H 2 SO 4 , 0.0231 mol PbSO 4 , (0.023056 × 2) = 0.0461 mol HC 2 H 3 O 2 are present after reaction 0.053405 mol H 2 SO 4 × 98.09 g/mol = 5.2385 = 5.24 g H 2 SO 4 0.023056 mol PbSO 4 × 303.3 g/mol = 6.9928 = 6.99 g PbSO 4 0.046111 mol HC 2 H 3 O 2 × 60.05 g/mol = 2.7690 = 2.77 g HC 2 H 3 O 2 Check. The initial mass of reactants was 15.00 g; and the final mass of excess reactant and products is 15.00 g; mass is conserved. 3.77 Analyze. Given: amounts of two reactants. Find: theoretical yield. Plan. Determine the limiting reactant and the maximum amount of product it could produce. Then calculate % yield. Solve. 64 3 Stoichiometry (a) 30.0 g C 6 H 6 × 65.0 g Br2 × 1 mol Br2 = 0.4068 = 0.407 mol Br2 159.8 g Br2 Solutions to Exercises 1 mol C 6 H 6 = 0.3841 = 0.384 mol C 6 H 6 78.11 g C 6 H 6 Since C 6 H 6 and Br 2 react in a 1:1 mole ratio, C 6 H 6 is the limiting reactant and determines the theoretical yield. 0.3841 mol C 6 H 6 × 1 mol C 6 H 5 Br 157.0 g C 6 H 5 Br × = 60.30 = 60.3 g C 6 H 5 Br 1 mol C 6 H 6 1 mol C 6 H 5 Br Check. 30/78 ~ 3/8 mol C 6 H 6 . 65/160 ~ 3/8 mol Br 2 . Since moles of the two reactants are similar, a precise calculation is needed to determine the limiting reactant. 3/8 × 160 ≈ 60 g product (b) 3.78 (a) % yield = 42.3 g C 6 H 5 Br actual 60.3 g C 6 H 5 Br theoretical × 100 = 70.149 = 70.10% C 2 H 6 + Cl 2 → C 2 H 5 Cl + HCl 125 g C 2 H 6 × 255 g Cl 2 × 1 mol C 2 H 6 = 4.157 = 4.16 mol C 2 H 6 30.07 g C 2 H 6 1 mol Cl 2 = 3.596 = 3.60 mol Cl 2 70.91 g Cl 2 Since the reactants combine in a 1:1 mole ratio, Cl 2 is the limiting reactant. The theoretical yield is: 3.596 mol Cl 2 × 1 mol C 2 H 5 Cl 64.51 g C 2 H 5 Cl × = 231.98 = 232 g C 2 H 5 Cl 1 mol Cl 2 1 mol C 2 H 5 Cl (b) 3.79 % yield = 206 g C 2 H 5 Cl actual × 100 = 88.8% 232 g C 2 H 5 Cl theoretical Analyze. Given: g of two reactants, % yield. Find: g S8. Plan. Determine limiting reactant and theoretical yield. Use definition of % yield to calculate actual yield. Solve. 30.0 g H 2 S × 50.0 g O 2 × 1 mol H 2 S = 0.8803 = 0.880 mol H 2 S 34.08 g H 2 S 1 mol O 2 = 1.5625 = 1.56 mol O 2 32.00 g H 2 S 4 mol O 2 = 0.4401 = 0.440 mol O 2 required 8 mol H 2 S 0.8803 mol H 2 S × Since there is more than enough O2 to react exactly with 0.880 mol H2S, O2 is present in excess and H2S is the limiting reactant. 0.8803 mol H 2 S × 256.56 g S 8 1 mol S 8 × = 28.231 = 28.2 g S 8 theoretical yield 8 mol H 2 S 1 mol S 8 65 3 Stoichiometry % yield = Solutions to Exercises Check. 30/34 ≈ 1 mol H2S; 50/32 ≈ 1.5 mol O 2 . Twice as many mol H2S as mol O2 are required, so H2S limits. 1 × (260/8) ≈ 30 g S8 theoretical. % yield × theoretical actual × 100; = actual yield theoretical 100 98% × 28.231 g S 8 = 27.666 = 28 g S 8 actual 100 3.80 H 2 S(g) + 2NaOH(aq) → Na 2 S(aq) + 2H 2 O(l) 1.50 g H 2 S × 1 mol H 2 S = 0.04401 = 0.0440 mol H 2 S 34.08 g H 2 S 1 mol NaOH 2.00 g NaOH × = 0.0500 mol NaOH 40.00 g NaOH By inspection, twice as many mol NaOH as H 2 S are needed for exact reaction, but mol NaOH given is less than twice mol H 2 S, so NaOH limits. 0.0500 mol NaOH × 1 mol Na 2 S 78.05 g Na 2 S × = 1.95125 = 1.95 g Na 2 S theoretical 2 mol NaOH 1 mol Na 2 S 92.0 % × 1.95125 g Na 2 S theoretical = 1.7951 = 1.80 g Na 2 S actual 100 Additional Exercises 3.81 (a) (b) (c) 3.82 CH3COOH = C2H4O2. At room temperature and pressure, pure acetic acid is a liquid. C2H4O2(l) + 2 O2(g) → 2 CO2(g) + 2 H2O(l) Ca(OH) 2 (s) → CaO(s) + H 2 O(g) Ni(s) + Cl 2 (g) → NiCl 2 (s) The formulas of the fertilizers are NH 3 , NH 4 NO 3 , (NH 4 ) 2 SO 4 and (NH 2 ) 2 CO. Qualitatively, the more heavy, non-nitrogen atoms in a molecule, the smaller the mass % of N. By inspection, the mass of NH 3 is dominated by N, so it will have the greatest % N, (NH 4 ) 2 SO 4 will have the least. In order of increasing % N: (NH 4 ) 2 SO 4 < NH 4 NO 3 < (NH 2 ) 2 CO < NH 3 . Check by calculation: (NH 4 ) 2 SO 4 : FW = 2(14.0) + 8(1.0) + 1(32.1) + 4(16.0) = 132.1 amu % N = [2(14.0)/132.1] × 100 = 21.2% NH 4 NO 3 : FW = 2(14.0) + 4(1.0) + 3(16.0) = 80.0 amu % N = [2(14.0)/80.0] × 100 = 35.0% (NH 2 ) 2 CO: FW = 2(14.0) + 4(1.0) = 1(12.0) + 1(16.0) = 60.0 amu % N = [2(14.0)/60.0] × 100 = 46.7% N NH 3 : FW = 1(14.0) + 3(1.0) = 17.0 % N = [14.0/17.0] × 100 = 82.4 % N 66 3 Stoichiometry 3.83 (a) 1.25 carat × 0.200 g 1 carat × 0.020816 mol C × Solutions to Exercises 1 mol C = 0.020816 = 0.0208 mol C 12.01 g C 6.022 × 10 23 C atoms = 1.25 × 10 22 C atoms 1 mol C 1 mol C 9 H 8 O 4 = 2.7747 × 10 − 3 = 2.77 × 10 − 3 mol HC 9 H 7 O 4 180.2 g C 9 H 8 O 4 6.022 × 10 23 molecules = 1.67 × 10 21 HC 9 H 7 O 4 molecules 1 mol (b) 0.500 g C 9 H 8 O 4 × 0.0027747 mol C 9 H 8 O 4 × 3.84 (a) (b) 5.342 × 10 −21 g 6.0221 × 10 23 molecules × = 3217 g/mol penicillin G 1 molecule penicillin G 1 mol 1.00 g hemoglobin (hem) contains 3.40 × 10 – 3 g Fe. 1.00 g hem 3.40 × 10 −3 g Fe × 55.85 g Fe 4 mol Fe × = 6.57 × 10 4 g/mol hemoglobin 1 mol Fe 1 mol hem 3.85 (a) Analyze. Given: diameter of Si sphere (dot), density of Si. Find: mass of dot. Plan. Calculate volume of sphere in cm3, use density to calculate mass of the sphere (dot). Solve. V = 4/3π r3; r = d/2 radius of dot = 4 nm 2 × 1 × 10 −9 m 1 nm × 1 cm 1 × 10 −9 m = 2 × 10 −7 cm volume of dot = (4/3) × π × (2 × 10– 7)3 = 3.35 × 10– 20 = 3 × 10– 20 cm3 3.35 × 10 − 20 cm 3 × 2.3 g Si cm 3 = 7.707 × 10 − 20 = 8 × 10 − 20 g Si in dot (b) Plan. Change g Si to mol Si using molar mass, then mol Si to atoms Si using Avogadro’s number. Solve. 7.707 × 10 − 20 g Si × 1 mol Si 28.0855 g Si × 6.022 × 10 23 Si atoms mol Si = 1.653 × 10 3 = 2 × 103 Si atoms (c) Plan. A 4 nm quantum dot of Ge also has a volume of 3 × 10– 20 cm3. Use density of Ge and Avogadro’s number to calculate the number of Ge atoms in a 4 nm spherical quantum dot. 3.35 × 10 − 20 cm 3 × 5.325 g Ge cm 3 × 1 mol Ge 72.64 g Ge × 6.022 × 10 23 Ge atoms mol Ge = 1.479 × 10 3 = 1 × 10 3 Ge atoms Strictly speaking, the result has 1 sig fig (from 4 nm). A more meaningful comparison might be 1700 Si atoms vs. 1500 Ge atoms. Although Ge has greater molar mass, it is also more than twice as dense as Si, so the numbers of atoms in the Si and Ge dots are similar. 67 3 Stoichiometry 3.86 68.2 g C × 6.86 g H × 15.9 g N × 9.08 g O × 1 mol C = 5.68 mol C; 5.68/0.568 ≈ 10 12.01 g C 1 mol H = 6.81 mol H; 6.81/0.568 ≈ 12 1.008 g H 1 mol N = 1.13 mol N; 1.13/0.568 ≈ 2 14.01 g N 1 mol O = 0.568 mol O; 0.568/0.568 = 1 16.00 g O Solutions to Exercises Plan. Assume 100 g, calculate mole ratios, empirical formula, then molecular formula from molar mass. Solve. The empirical formula is C 1 0H 1 2N 2 O, FW = 176 amu (or g). Since the molar mass is 176, the empirical and molecular formula are the same, C 1 0H 1 2N 2 O. 3.87 Plan. Assume 1.000 g and get mass O by subtraction. (a) 0.7787 g C × 0.1176 g H × 0.1037 g O × 1 mol C = 0.06484 mol C 12.01 g C 1 mol H = 0.1167 mol H 1.008 g H 1 mol C = 0.006481 mol O 16.00 g O Solve. Dividing through by the smallest of these values we obtain C 1 0H 1 8O. (b) The formula weight of C 1 0H 1 8O is 154. Thus, the empirical formula is also the molecular formula. 3.88 Since all the C in the vanillin must be present in the CO 2 produced, get g C from g CO 2 . 2.43 g CO 2 × 1 mol CO 2 12.01 g C × = 0.6631 = 0.663 g C 44.01 g CO 2 1 mol C Since all the H in vanillin must be present in the H 2 O produced, get g H from g H 2 O. 0.50 g H 2 O × 1.008 g H 1 mol H 2 O 2 mol H × × = 0.0559 = 0.056 g H 18.02 g H 2 O 1 mol H 2 O 1 mol H Get g O by subtraction. (Since the analysis was performed by combustion, an unspecified amount of O 2 was a reactant, and thus not all the O in the CO 2 and H 2 O produced came from vanillin.) 1.05 g vanillin – 0.663 g C – 0.056 g H = 0.331 g O 0.6631 g C × 0.0559 g H × 1 mol C = 0.0552 mol C; 0.0552 / 0.0207 = 2.67 12.01 g C 1 mol H = 0.0555 mol C; 0.0555 / 0.0207 = 2.68 1.008 g H 68 3 Stoichiometry 0.331 g O × Solutions to Exercises 1 mol O = 0.0207 mol O; 0.0207 / 0.0207 = 1.00 16.00 g O Multiplying the numbers above by 3 to obtain an integer ratio of moles, the empirical formula of vanillin is C 8 H 8 O 3 . 3.89 Plan. Because different sample sizes were used to analyze the different elements, calculate mass % of each element in the sample. i. ii. iii. iv. Solve. i. 3.52 g CO 2 × 12.01 g C 1 mol CO 2 1 mol C × × = 0.9606 = 0.961 g C 44.01 g CO 2 1 mol CO 2 1 mol C Calculate mass % C from g CO 2 . Calculate mass % Cl from AgCl. Get mass % H by subtraction. Calculate mole ratios and the empirical formulas. 0.9606 g C × 100 = 64.04 = 64.0% C 1.50 g sample ii. 1.27 g AgCl × 1 mol AgCl 35.45 g Cl 1 mol Cl × × = 0.3142 = 0.314 g Cl 143.3 g AgCl 1 mol AgCl 1 mol Cl 0.3142 g Cl × 100 = 31.42 = 31.4% Cl 1.00 g sample iii. iv. % H = 100.0 – (64.04% C + 31.42% Cl) = 4.54 = 4.5% H Assume 100 g sample. 64.04 g C × 1 mol C = 5.33 mol C; 5.33 / 0.886 = 6.02 12.01 g C 1 mol Cl = 0.886 mol Cl; 0.886 / 0.886 = 1.00 35.45 g Cl 31.42 g Cl × 4.54 g H × 1 mol H = 4.50 mol H; 4.50 / 0.886 = 5.08 1.008 g H The empirical formula is probably C 6 H 5 Cl. The subscript for H, 5.08, is relatively far from 5.00, but C 6 H 5 Cl makes chemical sense. More significant figures in the mass data are required for a more accurate mole ratio. 3.90 The mass percentage is determined by the relative number of atoms of the element times the atomic weight, divided by the total formula mass. Thus, the mass percent of 79.91 bromine in KBrO x is given by 0.5292 = . Solving for x, we obtain 39.10 + 79.91 + x(16.00) x = 2.00. Thus, the formula is KBrO 2 . (a) Let AW = the atomic weight of X. According to the chemical reaction, moles XI 3 reacted = moles XCl 3 produced 3.91 69 3 Stoichiometry 0.5000 g XI 3 × 1 mol XI 3 / (AW + 380.71) g XI 3 Solutions to Exercises = 0.2360 g XCl 3 × 1 mol XCl 3 ( AW + 106.36) g XCl 3 0.5000 (AW + 106.36) = 0.2360 (AW + 380.71) 0.5000 AW + 53.180 = 0.2360 AW + 89.848 0.2640 AW = 36.67; AW = 138.9 g (b) 3.92 X is lanthanum, La, atomic number 57. C 2 H 5 OH(l) + 3O 2 (g) → 2CO 2 (g) + 3H 2 O(g) C 3 H 8 (g) + 5O 2 (g) → 3CO 2 (g) + 4H 2 O(g) CH 3 CH 2 COCH 3 (l) + 11/2 O 2 (g) → 4CO 2 (g) + 4H 2 O(l) In a combustion reaction, all H in the fuel is transformed to H 2 O in the products. The reactant with most mol H/mol fuel will produce the most H 2 O. C 3 H 8 and CH 3 CH 2 COCH 3 (C 4 H 8 O) both have 8 mol H/mol fuel, so 1.5 mol of either fuel will produce the same amount of H 2 O. 1.5 mol C 2 H 5 OH will produce less H 2 O. 3.93 O 3 (g) + 2NaI(aq) + H 2 O(l) → O 2 (g) + I 2 (s) + 2NaOH(aq) (a) 5.95 × 10 −6 mol O 3 × 1.3 mg O 3 × 2 mol NaI = 1.19 × 10 − 5 mol NaI 1 mol O 3 × 1 mol O 3 2 mol NaI 149.9 g NaI × × 48.00 g O 3 1 mol O 3 1 mol NaI (b) 1 × 10 −3 g 1 mg = 8.120 × 10 – 3 = 8.1 × 10 – 3 g NaI = 8.1 mg NaI 3.94 2NaCl(aq) + 2H 2 O(l) → 2NaOH(aq) + H 2 (g) + Cl 2 (g) Calculate mol Cl 2 and relate to mol H 2 , mol NaOH. 1.5 × 10 6 kg × 1000 g 1 mol Cl 2 × = 2.115 × 10 7 = 2.1 × 10 7 mol Cl 2 1 kg 70.91 g Cl 2 1 mol H 2 2.016 g H 2 × = 4.26 × 10 7 g H 2 = 4.3 × 10 4 kg H 2 1 mol Cl 2 1 mol H 2 2.115 × 10 7 mol Cl 2 × 4.3 × 10 7 g × 1 metric ton = 43 metric tons H 2 1 × 10 6 g (1 Mg) 2 mol NaOH 40.0 g NaOH × = 1.69 × 10 9 = 1.7 × 10 9 g NaOH 1 mol Cl 2 1 mol NaOH 2.115 × 10 7 mol Cl 2 × 1.7 × 10 9 g NaOH = 1.7 × 10 6 kg NaOH = 1.7 × 10 3 metric tons NaOH 3.95 2C 5 7H 1 10O 6 + 163O 2 → 114CO 2 + 110H 2 O molar mass of fat = 57(12.01) + 110(1.008) + 6(16.00) = 891.5 1.0 kg fat × 1000 g 1 mol fat 110 mol H 2 O 18.02 g H 2 O 1 kg × × × × = 1.1 kg H 2 O 1 kg 891.5 g fat 2 mol fat 1 mol H 2 O 1000 g 70 3 Stoichiometry 3.96 (a) 0.467 g CO × Solutions to Exercises Plan. Calculate the total mass of C from g CO and g CO 2 . Calculate the mass of H from g H 2 O. Calculate mole ratios and the empirical formula. Solve. 1 mol CO 1 mol C × × 12.01 g C = 0.200 g C 28.01 g CO 1 mol CO 1 mol CO 2 1 mol C × × 12.01 g C = 0.200 g C 44.01 g CO 2 1 mol CO 2 0.733 g CO 2 × Total mass C is 0.200 g + 0.200 g = 0.400 g C. 0.450 g H 2 O × 1.008 g H 1 mol H 2 O 2 mol H × × = 0.0503 g H 18.02 g H 2 O 1 mol H 2 O 1 mol H (Since hydrocarbons contain only the elements C and H, g H can also be obtained by subtraction: 0.450 g sample – 0.400 g C = 0.050 g H.) 0.400 g C × 1 mol C = 0.0333 mol C; 0.0333 / 0.0333 = 1.0 12.01 g C 1 mol H = 0.0499 mol H; 0.0499 / 0.0333 = 1.5 1.008 g H 0.0503 g H × Multiplying by a factor of 2, the empirical formula is C 2 H 3 . (b) Mass is conserved. Total mass products – mass sample = mass O 2 consumed. 0.467 g CO + 0.733 g CO 2 + 0.450 g H 2 O – 0.450 g sample = 1.200 g O 2 consumed (c) For complete combustion, 0.467 g CO must be converted to CO 2 . 2CO(g) + O 2 (g) → 2CO 2 (g) 0.467 g CO × 1 mol CO 1 mol O 2 32.00 g O 2 × × = 0.267 g O 2 28.01 g C 2 mol CO 1 mol O 2 The total mass of O 2 required for complete combustion is 1.200 g + 0.267 g = 1.467 g O 2 . 3.97 N 2 (g) + 3H 2 (g) → 2NH 3 (g) Determine the moles of N 2 and H 2 required to form the 3.0 moles of NH 3 present after the reaction has stopped. 3.0 mol NH 3 × 3.0 mol NH 3 × 3 mol H 2 = 4.5 mol H 2 reacted 2 mol NH 3 1 mol N 2 = 1.5 mol N 2 reacted 2 mol NH 3 mol H 2 initial = 3.0 mol H 2 remain + 4.5 mol H 2 reacted = 7.5 mol H 2 mol N 2 initial = 3.0 mol N 2 remain + 1.5 mol N 2 reacted = 4.5 mol N 2 71 3 Stoichiometry In tabular form: initial reaction final N 2 (g) 4.5 mol –1.5 mol 3.0 mol + 3H 2 (g) 7.5 mol –4.5 mol 3.0 mol Solutions to Exercises → 0 2NH 3 (g) mol +3.0 mol 3.0 mol (Tables like this will be extremely useful for solving chemical equilibrium problems in Chapter 15.) 3.98 All of the O 2 is produced from KClO 3 ; get g KClO 3 from g O 2 . All of the H 2 O is produced from KHCO 3 ; get g KHCO 3 from g H 2 O. The g H 2 O produced also reveals the g CO 2 from the decomposition of NaHCO 3 . The remaining CO 2 (13.2 g CO 2 – g CO 2 from NaHCO 3 ) is due to K 2 CO 3 and g K 2 CO 3 can be derived from it. 4.00 g O 2 × 1 mol O 2 2 mol KClO 3 122.6 g KClO 3 × × = 10.22 = 10.2 g KClO 3 32.00 g O 2 3 mol O 2 1 mol KClO 3 1 mol H 2 O 2 mol KHCO 3 100.1 g KHCO 3 × × = 20.00 = 20.0 g KHCO 3 18.02 g H 2 O 1 mol H 2 O 1 mol KHCO 3 1 mol H 2 O 2 mol CO 2 44.01 g CO 2 × × = 8.792 = 8.79 g CO 2 from KHCO 3 18.02 g H 2 O 1 mol H 2 O 1 mol CO 2 1.80 g H 2 O × 1.80 g H 2 O × 13.20 g CO 2 total – 8.792 CO 2 from KHCO 3 = 4.408 = 4.41 g CO 2 from K 2 CO 3 4.408 g CO 2 × 1 mol CO 2 1 mol K 2 CO 3 138.2 g K 2 CO 3 × × = 13.84 = 13.8 g K 2 CO 3 44.01 g CO 2 1 mol CO 2 1 mol K 2 CO 3 100.0 g mixture – 10.22 g KClO 3 – 20.00 g KHCO 3 – 13.84 g K 2 CO 3 = 56.0 g KCl 3.99 (a) (b) 2C 2 H 2 (g) + 5O 2 (g) → 4CO 2 (g) + 2H 2 O(g) Following the approach in Sample Exercise 3.18, 10.0 g C 2 H 2 × 32.00 g O 2 1 mol C 2 H 2 5 mol O 2 × × = 30.7 g O 2 required 26.04 g C 2 H 2 2 mol C 2 H 2 1 mol O 2 Only 10.0 g O 2 are available, so O 2 limits. (c) Since O 2 limits, 0.0 g O 2 remain. Next, calculate the g C 2 H 2 consumed and the amounts of CO 2 and H 2 O produced by reaction of 10.0 g O 2 . 10.0 g O 2 × 1 mol O 2 2 mol C 2 H 2 26.04 g C 2 H 2 × × = 3.26 g C 2 H 2 consumed 32.00 g O 2 5 mol O 2 1 mol C 2 H 2 10.0 g C 2 H 2 initial – 3.26 g consumed = 6.74 = 6.7 g C 2 H 2 remain 10.0 g O 2 × 10.0 g O 2 × 1 mol O 2 4 mol CO 2 44.01 g CO 2 × × = 11.0 g CO 2 produced 32.00 g O 2 5 mol O 2 1 mol CO 2 1 mol O 2 2 mol H 2 O 18.02 g H 2 O × × = 2.25 g H 2 O produced 32.00 g O 2 5 mol O 2 1 mol H 2 O 72 3 Stoichiometry 3.100 (a) 1.5 × 10 5 g C 9 H 8 O 4 × Solutions to Exercises 1 mol C 9 H 8 O 4 1 mol C 7 H 6 O 3 138.1 g C 7 H 6 O 3 × × 180.2 g C 9 H 8 O 4 1 mol C 9 H 8 O 4 1 mol C 7 H 6 O 3 = 1.1496 × 105 g = 1.1 × 102 kg C7H6O3 (b) (c) If only 80 percent of the acid reacts, then we need 1/0.80 = 1.25 times as much to obtain the same mass of product: 1.25 × 1.15 × 10 2 kg = 1.4 × 10 2 kg C 7 H 6 O 3 Calculate the number of moles of each reactant: 1.85 × 10 5 g C 7 H 6 O 3 × 1.25 × 10 5 g C 4 H 6 O 3 × 1 mol C 7 H 6 O 3 = 1.340 × 10 3 = 1.34 × 10 3 mol C 7 H 6 O 3 138.1 g C 7 H 6 O 3 1 mol C 4 H 6 O 3 = 1.224 × 10 3 = 1.22 × 10 3 mol C 4 H 6 O 3 102.1 g C 4 H 6 O 3 We see that C 4 H 6 O 3 limits, because equal numbers of moles of the two reactants are consumed in the reaction. 1.224 × 10 3 mol C 4 H 6 O 3 × 1 mol C 9 H 8 O 4 180.2 g C 9 H 8 O 4 × = 2.206 × 10 5 1 mol C 7 H 6 O 3 1 mol C 9 H 8 O 4 = 2.21 × 10 5 gC 9 H 8 O 4 (d) percent yield = 1.82 × 10 5 g 2.206 × 10 5 g × 100 = 82.5% Integrative Exercises 3.101 density Plan. Volume cube ⎯⎯ ⎯ ⎯ → mass CaCO 3 → moles CaCO 3 → moles O → O atoms ⎯ Solve. (2.005) 3 in 3 × ( 2.54) 3 cm 3 1 in 3 × 2.71 g CaCO 3 1 cm 3 × 1 mol CaCO 3 3 mol O × 100.1 g CaCO 3 1 mol CaCO 3 × 6.022 × 10 23 O atoms = 6.46 × 10 24 O atoms 1 mol O 3.102 (a) density Plan. volume of Ag cube ⎯⎯ ⎯ ⎯ → mass of Ag → mol Ag → Ag atoms ⎯ Solve. (1.000) 3 cm 3 Ag × 10.5 g Ag 1 cm 3 Ag 107.87 g Ag × 1 mol Ag × 6.022 × 10 23 atoms 1 mol = 5.8618 × 10 22 = 5.86 × 10 22 Ag atoms (b) 1.000 cm 3 cube volume, 74% is occupied by Ag atoms 0.74 cm 3 = volume of 5.86 × 10 2 2 Ag atoms 0.7400 cm 3 5.8618 × 10 22 Ag atoms = 1.2624 × 10 − 23 = 1.3 × 10 − 23 cm 3 / Ag atom Since atomic dimensions are usually given in Å, we will show this conversion. 73 3 Stoichiometry 1.2624 × 10 − 23 cm 3 × (1 × 10 −2 ) 3 m 3 1 cm 3 × 1Å3 Solutions to Exercises (1 × 10 −10 ) 3 m 3 = 12.62 = 13 Å 3 / Ag atom (c) V = 4/3 π r 3 ; r 3 = 3V/4π; r = (3V/4π) 1 /3 r A = (3 × 12.62 Å 3 / 4π) 1 /3 = 1.444 = 1.4 Å 3.103 (a) Analyze. Given: gasoline = C 8 H 1 8, density = 0.69 g/mL, 20.5 mi/gal, 225 mi. Find: kg CO 2 . Plan. Write and balance the equation for the combustion of octane. Change mi → gal octane → mL → g octane. Use stoichiometry to calculate g and kg CO 2 from g octane. Solve. 2C 8 H 1 8(l) + 25O 2 (g) → 16CO 2 (g) + 18H 2 O(l) 225 mi × 1 gal 0.69 g octane 3.7854 L 1 mL × × × = 2.8667 × 10 4 g −3 20.5 mi 1 gal 1 mL 1 × 10 L = 29 kg octane 2.8667 × 10 4 g C 8 H 18 × 1 mol C 8 H 18 16 mol CO 2 44.01 g CO 2 × × = 8.8382 × 10 4 g 114.2 g C 8 H 18 2 mol C 8 H 18 1 mol CO 2 = 88 kg CO 2 ⎛ 225 × 4 × 0.7 ⎞ 3 3 3 Check. ⎜ ⎟ × 10 = ( 45 × 0.7 ) × 10 = 30 × 10 g = 30 kg octane 20 ⎝ ⎠ 44 1 30 kg × 8 ≈; ≈ 80 kg CO 2 114 3 3 (b) Plan. Use the same strategy as part (a). Solve. 225 mi × 1 gal 5 mi × 0.69 g octane 3.7854 L 1 mL × × = 1.1754 × 10 5 −3 1 gal 1 mL 1 × 10 L = 1 × 10 2 kg octane 1.1754 × 10 5 g C 8 H 18 × 1 mol C 8 H 18 16 mol CO 2 44.01 g CO 2 × × = 3.624 × 10 5 g 114.2 g C 8 H 18 2 mol C 8 H 18 1 mol CO 2 = 4 × 10 2 kg CO 2 Check. Mileage of 5 mi/gal requires ~4 times as much gasoline as mileage of 20.5 mi/gal, so it should produce ~4 times as much CO2. 90 kg CO2 [from (a)] × 4 = 360 = 4 × 102 kg CO2 [from (b)]. 3.104 Plan. We can proceed by writing the ratio of masses of Ag to AgNO 3 , where y is the atomic mass of nitrogen. Solve. Ag 107.8682 = 0.634985 = AgNO 3 107.8682 + 3(15.9994) + y Solve for y to obtain y = 14.0088. This is to be compared with the currently accepted value of 14.0067. 74 3 Stoichiometry 3.105 (a) (b) Solutions to Exercises S(s) + O 2 (g) → SO 2 (g); SO 2 (g) + CaO(s) → CaSO 3 (s) If the coal contains 2.5% S, then 1 g coal contains 0.025 g S. 1 kg 1000 g 0.025 g S 1 mol S 2000 tons coal 2000 lb × × × × × day 1 ton 2.20 lb 1 kg 1 g coal 32.07 g S × 1 kg CaO 1 mol SO 2 1 mol CaO 56.08 g CaO × × × = 1 mol S 1 mol SO 2 1 mol CaO 1000 g CaO 79, 485 = 7.9 × 10 4 kg CaO or 7.9 × 10 7 g CaO (c) 1 mol CaO = 1 mol CaSO3 7.9485 × 107 g CaO × 1 mol CaO 1 mol CaSO 3 120.14 g CaSO 3 × × 56.08 g CaO 1 mol CaO 1 mol CaSO 3 = 1.703 × 10 8 = 1.7 × 10 8 g CaSO 3 This corresponds to about 190 tons of CaSO3 per day as a waste product. 3.106 Analyze. Given: 2.0 in × 3.0 in boards, 5000 boards, 0.65 mm thick Cu; 8.96 g/cm 3 Cu; 85% Cu removed; 97% yield for reaction. Find: mass Cu(NH 3 ) 4 Cl 2 , mass NH 3 . Plan. vol Cu/board × density → mass Cu/board → 5000 boards × 85% = total Cu removed = actual yield; actual yield/0.97 = theoretical yield Cu. mass Cu → mol Cu → mol Cu(NH 3 ) 4 Cl or NH 3 → desired masses. Solve. 2.0 in × 3.0 in × (2.54) 2 cm 2 1 cm × 0.65mm × = 2.516 = 2.5 cm 3 Cu/board 2 10 mm in 2.516 cm 3 Cu 8.96 g × × 5000 boards × 0.85 removed = 95,814 g = 96 kg Cu removed board cm 3 95 ,814 g Cu actual yield = 98 ,777 g = 99 kg Cu theoretical 0.97 98 ,777 g Cu × 202.575 g 1 mol Cu(NH 3 ) 4 Cl 2 1 mol Cu × × = 314 ,887 g 63.546 g Cu 1 mol Cu 1 mol Cu(NH 3 ) 4 Cl 2 = 3.1 × 10 2 kg Cu(NH 3 ) 4 Cl 2 98 ,777 g Cu × 4 mol NH 3 17.03 g NH 3 1 mol Cu × × = 105 , 891 g = 1.1 × 10 2 kg NH 3 63.546 g Cu 1 mol Cu mol NH 3 3.107 (a) Plan. Calculate the kg of air in the room and then the mass of HCN required to produce a dose of 300 mg HCN/kg air. Solve. 12 ft × 15 ft × 8.0 ft = 1440 = 1.4 × 10 3 ft 3 of air in the room 1440 ft 3 air × (12 in) 3 ( 2.54 cm) 3 0.00118 g air 1 kg × × × = 48.12 = 48 kg air 1000 g 1 ft 3 1 in 3 1 cm 3 air 48.12 kg air × 300 mg HCN 1g × = 14.43 = 14 g HCN 1 kg air 1000 mg 75 3 Stoichiometry (b) Solutions to Exercises 2NaCN(s) + H 2 SO 4 (aq) → Na 2 SO 4 (aq) + 2HCN(g) The question can be restated as: What mass of NaCN is required to produce 14 g of HCN according to the above reaction? 14.43 g HCN × 1 mol HCN 2 mol NaCN 49.01 g NaCN × × = 26.2 = 26 g NaCN 27.03 g HCN 2 mol HCN 1 mol NaCN × 30 oz 1 lb 454 g × × = 17 ,025 1 yd 2 16 oz 1 lb = 1.7 × 10 4 g acrilan in the room (c) 12 ft × 15 ft × 1 yd 2 9 ft 2 50% of the carpet burns, so the starting amount of CH 2 CHCN is 0.50(17,025) = 8,513 = 8.5 × 10 3 g 8 ,513 g CH 2 CHCN × 50.9 g HCN 100 g CH 2 CHCH = 4333 = 4.3 × 10 3 g HCN possible If the actual yield of combustion is 20%, actual g HCN = 4,333(0.20) = 866.6 = 8.7 × 10 2 g HCN produced. From part (a), 14 g of HCN is a lethal dose. The fire produces much more than a lethal dose of HCN. 3.108 (a) (b) N2(g) + O2(g) → 2ΝΟ(g); 2ΝΟ(g) + Ο2(g) → 2NO2(g) 1 million = 1 × 106 19 × 10 6 tons NO 2 × 2000 lb 453.6 g × = 1.724 × 10 13 = 1.7 × 10 13 g NO 1 ton 1 lb (c) Plan. Calculate g O2 needed to burn 500 g octane. This is 85% of total O2 in the engine. 15% of total O2 is used to produce NO2, according to the second equation in part (a). Solve. 2C8H18(l) + 25O2(g) → 16CO2(g) + 18H2O(l) 500 g C 8 H 18 × 1751 g O 2 total g O 2 32.00 g O 2 1 mol C 8 H 18 25 mol O 2 × × = 1751 = 1.75 × 10 3 g O 2 mol O 2 114.2 g C 8 H 18 2 mol C 8 H 18 = 0.85; 2060 = 2.1 × 10 3 g O 2 total in engine 2060 g O2 total × 0.15 = 309.1 = 3.1 × 109 g O2 used to produce NO2. One mol O2 produces 2 mol NO. Then 2 mol NO react with a second mol O2 to produce 2 mol NO2. Two mol O2 are required to produce 2 mol NO2; one mol O2 per mol NO2. 309.1 g O 2 × 1 mol O 2 32.00 g × 1 mol NO 2 1 mol O 2 × 46.01 g NO 2 1 mol O 2 = 444.4 = 4.4 × 10 2 g NO 2 76 ...
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This note was uploaded on 04/04/2009 for the course CHEM 121 taught by Professor Wyzlouzil during the Spring '07 term at Ohio State.

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