chm2045ch4 - Chemical Equations and Stoichiometry Chapter 4...

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Unformatted text preview: Chemical Equations and Stoichiometry Chapter 4 Reactants: Zn + I2 Product: ZnI2 Chapter goals Balance chemical equations. Perform stoichiometric calculations using balanced chemical equations. Limiting reagent. Calculate the theoretical and percent yields of a chemical reaction. Use stoichiometry to analyze a mixture of compounds or to determine the formula of a compound. Symbolism Reactants on left, products on right reactants separated from products by or reactants separated from each other with +, same for products heat h light balanced by both mass and charge by 2 using coefficients ; 2 H O use correct molecular formulas Symbolism (contd...) Subscripts appear after atoms (g) gas (l ) liquid (s) solid (aq) aqueous solution catalysts, special solvents, special conditions written above and/or below arrow Types of Chemical Reactions combination A + B C decomposition A B + C combustion A + O2(g) products + ...... Single replacement: AB + C A + CB Double replacement AB + CD AD + CB Balancing Chemical Equations done by inspection usually best to begin with largest molecule for combustion reactions involving organic compounds, balance C atoms first, then H atoms, and O last. convert any fractional coefficient to whole number by multiplying entire equation by the denominator of fraction Example Al(s) + O2(g) Al2O3(s) 2 Al(s) + O2(g) Al2O3(s) 2 Al(s) + 1.5 O2(g) Al2O3(s) fractional coefficients are not allowed then, multiply 2 (2 1.5 = 3) 4 Al(s) + 3 O2(g) 2 Al2O3(s) Example NH3(g) + O2(g) NO2(g) + H2O(l) 2 NH3(g) + O2(g) NO2(g) + 3 H2O(l) 2 NH3(g) + O2(g) 2 NO2(g) + 3 H2O(l) 3 .5 2 = 7 (4 + 3 ) 2 NH3(g) + 3.5 O2(g) 2 NO2(g) + 3 H2O(l) 4 NH3(g) + 7 O2(g) 4 NO2(g) + 6 H2O(l) Combustion Reactions reactions with molecular oxygen Hydrocarbons + molecular oxygen carbon dioxide and water e.g.. CH4(g) + O2(g) CO2(g) + H2O(g) (methane) CH4(g) + 2 O2(g) CO2(g) + 2H2O(g) Combustion Reactions C5H12(g) + O2(g) CO2(g) + H2O(l ) Balance C atoms C5H12(g) + O2(g) 5CO2(g) + H2O(l ) Balance H atoms C5H12(g) + O2(g) 5CO2(g) + 6H2O(l ) Balance O atoms, 5x2 + 6 = 16 O C5H12(g) + 8O2(g) 5CO2(g) + 6H2O(l ) Combustion Reactions C6H14(g) + O2(g) CO2(g) + H2O(g) Balance C atoms C6H14(g) + O2(g) 6CO2(g) + H2O(g) Balance H atoms C6H14(g) + O2(g) 6CO2(g) + 7H2O(g) Balance O atoms, 6x2 + 7 = 19 O C6H14(g) + 9.5O2(g) 6CO2(g) + 7H2O(g) Multiply entire equation by 2 2C6H14(g) + 19O2(g) 12CO2(g) + Stoichiometry calculations involving chemical equations Steps write a balanced chemical equation convert mass to moles determine limiting reagent if necessary calculate number of moles of desired substance (reactant or product) convert to desired units Example: Iron reacts with steam to form Fe O and H . Calculate mol H produced by reaction of 10.0 g iron with excess steam. Balanced eqn: 3 Fe(s) + 4 H2O(g) Fe3O4(s) + 4 H2(g) Next, convert g of Fe into moles Fe 1 mol Fe 10 gFe = 0.179 molFe 55.85 g Fe 4 mol H2 0.179 mol Fe = 0.239 mol H2 3 mol Fe Then, use mol Fe to calculate mol H . Water is not used because it is in excess Example: Tin(IV) oxide reacts with carbon at high temperature to form elemental tin and carbon monoxide. How many kg of carbon are needed to convert 50.0 kg of tin(IV) oxide to elemental tin? balanced equation: SnO2 + 2C Sn + 2CO SnO + 2 C Sn + 2 CO 50.0 kg kg ? 1000 g 1 mol SnO 2 50.0 kg SnO 2 = 331.8 mol SnO 2 1 kg 150.7 g SnO 2 2 2 mol C 12.01 g C 1 kg C 331.8 mol SnO 2 1 mol SnO 2 1 mol C 1000 g C convert to mol C gC kg C ANS = 7.97 kg C 1.40 mol O , 7.00 mol N , with traces of other gases. A small lamp fueled by methanol, CH OH, is lighted in the container. How many mL of methanol (d = 0.791 g/mL) will balanced equation: lamp goes out? be consumed when the 2CH3OH(l) + 3O2(g) 2CO2(g) + 4H2O(g) 2 mol CH3OH 1.40 mol O 2 = 0.933 mol CH 3 OH 3 mol O 2 Determine mol of CH3OH from mol of O2 2 CH OH(l) + 3 O (g) 2 CO (g) + 4 H O(l) 32.04 g CH 3 OH mL 0.933 mol CH 3 OH 1 mol CH3 OH 0.791 g 3 density = 37.8 mL CH3OH Limiting Reagent (Reactant) the reactant present in the exact quantity to react completely with other reactant; -consumed completely in the reaction; -determines amount of product yielded. excess reagent (reactant): the reactant present in greater quantity in the reaction; -some of it remains after the reaction is completed 4 Li(s) start (given) 4.0 mol reacted -4.0 mol end 0 mol + O2(g) 2 Li2O(s) 1.0 mol 0 mol -1.0 mol +2.0 mol 0 mol 2.0 mol NoLR For the 4.0 mol of Li to react we need (? mol of O2) 1 mol O 2 4.0 mol Li = 1 mol O 2 needed 4 mol Li Then, there is no L. R (ratio according to equation). 2 mol Li 2O 4.0 mol Li = 2 mol Li2 O produced 4 mol Li 4 Li(s) start (given) 4.0 mol reacted -2.0 mol + O2(g) 2 Li2O(s) 0.5 mol 0 mol -0.5 mol +1.0 mol 4 mol Li 0.5 mol O 2 = 2 mol Li needed 1 mol O 2 2 mol Li2 O 0.5 mol O 2 = 1 mol Li2 O produced 1 mol O 2 Limiting/excess reactant 4 Li(s) start 8.0 mol reacted -8.0 mol end 0 mol 2 + O2(g) 3.0 mol -2.0 mol 1.0 mol xs 2 Li2O(s) 0 mol +4.0 mol 4.0 mol Li LR 1 mol O 2 8 = 2 mol O 2 needed6.0 mol If O ? mol Li mol 12.0 -3.0 mol 4 mol Li 2 mol Li2 O 8 mol Li = 4 mol Li2 O produced 4 mol Li Example: Copper(II) oxide reacts with ammonia to form copper, water, and nitrogen. If 236.1 g copper(II) oxide are treated with 64.38 g ammonia, how much copper (grams) is produced? How many g of excess reagent remain? Balanced equation 3CuO + 2NH3(g) 3Cu(s) + 3H2O(l) + N2(g) 236.1 g 64.38 g ?g 3CuO + 2NH (g) 3Cu(s) + 3H O(l) + N (g) 1 mol CuO 236.1 g CuO = 2.968 mol CuO 79.55 g CuO 1 mol NH 3 64.38 g NH 3 = 3.780 mol NH 3 17.03 g NH 3 How many moles of NH3 are needed for all CuO to react completely? 2 mol NH 3 2.968 mol CuO = 1.979 mol NH 3 3 mol CuO Given 3.780 mol NH3 (in excess), i.e., CuO is the LR 3CuO + 2NH (g) 3Cu(s) + 3H O(l) + N (g) We can also ask how many moles of CuO are needed for all NH3 to react completely? 3 mol CuO 3.780 mol NH 3 = 5.670 mol CuO 2 mol NH 3 But we do not have 5.670 mol CuO!!! Only 2.968 mol CuO. So CuO must be the LR!!! 3.780 mol NH 3 2.968 mol CuO = 0.989 < = 1.890 3 mol CuO 2 mol NH 3 Conclusion: we must use the amount of CuO (2.968 mol) to calculate the amount of Cu obtained in the reaction. 3CuO + 2NH (g) 3Cu(s) + 3H O(l) + N (g) How many moles of Cu are obtained? 3 mol Cu 2.968 mol CuO = 2.968 mol Cu 3 mol CuO 63.55 g Cu 2.968 mol CuO = 188.6 g Cu 1 mol Cu 17.03 g NH 3 1.801 mol NH 3 = 30.67 g NH 3 1 mol NH 3 Example: Dihydrogen sulfide and sulfur dioxide react to form sulfur & water. How much sulfur is formed when 5.00 g dihydrogen sulfide are mixed with 5.00 g sulfur dioxide? How many g of excess reagent remain after reaction? Equation 16 H2S(g) + 8 SO2(g) 3 S8(s) + 16H2O(l) Firstly: calculate moles (of molecules) of H S and SO 1 mol H2S 5.00 g H2S x ---------- = 0.147 mol H2S 34.1 g H2S 1 mol SO2 5.00 g SO2 x ---------- = 0.0780 mol SO2 64.1 g SO2 Now, find out which is the limiting reagent 16 H2S(g) + 8 SO2(g) 3 S8(s) + 16 H2O(l) How many moles of H2S are needed for all SO2 to react? 16 mol H2S more 0.0780 mol SO2 x ---------- = 0.156 mol H2S than 8 mol SO2 we have We need 0.156 mol of H S to react with 0.0780 mol SO , but we have only 0.147 mol H S. This means that H S is the Limiting Reactant. Then, SO2 is the excess reagent. To determine the amount of sulfur we use H2S: 16 H2S(g) + 8 SO2(g) 3 S8(s) + 16 H2O(l) 3 mol S8 0.147 mol H2S x ---------- = 0.0276 mol S8 16 mol H2S MW (S ) = 8 x 32.066 = 256.6 g/mol 256.5 g S8 0.0276 mol S8 x ---------- = 7.08 g S8 1 mol S8 Alternatively....... Calc. moles of S8 produced by both reactants. 16 H2S(g) + 8 SO2(g) 3 S8(s) + 16 H2O(l) The reactant producing the smaller # mol S8 is the LIMITING REATANT. 3 mol S 8 0.0780 mol SO 2 = 0.0292 mol S 8 8 mol SO 2 3 mol S 8 0.147 mol H2 S = 0.0276 mol S 8 16 mol H2 S 0.147 mol H2S produces the smaller # mol S8 ; H2S is the LIMITING REATANT. 256.5 g S8 0.0276 mol S8 x ---------- = 7.08 g S8 1 mol S We need to determine the amount of SO2 that reacts reaction? This is (the LR): SO with 0.147 mol of H2S 16 H2S(g) + 8 SO2(g) 3 S8(s) + 16 H2O(l) 8 mol SO2 0.147 mol H2S x ---------- = 0.0735 mol SO2 16 mol H2S 0.0780 mol SO2 initial 0.0735 mol reacted = 0.0045 mol remaining after reaction If you begin with 2.70 g of Al and 4.05 g of Cl2, a) Which reactant is limiting? 0.100 mol Al 1 mol Al 2 mol AlCl3 2.70 g Al = 0.100 mol AlCl3 27.0 g Al 2 mol Al 1 mol Cl2 2 mol AlCl3 4.05 g Cl2 = 0.0381 mol AlCl3 70.9 g Cl2 3 mol Cl2 2 3 2 b) What mass of AlCl3 can be produced? b) What mass of the excess reactant remains when the reaction is finished? That is Al. The remaining Al will be the initial minus the reacted. For calculating the reacted we must use Cl2, the LR. 2 mol Al 0.0571 mol Cl2 = 0.0381 mol Al (reacted) 3 mol Cl2 0.100 mol (initial) - 0.0381 mol (reacted) = 0.0619 mol remaining 27.0 g Al 0.0619 mol Al = 1.67 g Al 1 mol Al Percent Yield Theoretical Yield: the maximum amount of product that can be obtained from a chemical reaction. It is the one calculated from the chemical equations Actual Yield: the amount of product that is obtained from a reaction--is less than the theoretical yield (due to ...) actual yield % yield = ---------------- x 100 theoretical yield % yield 100% 64.0 g of methanol is the theoretical yield (expected) 56.0 g is the actual yield (in the laboratory) 56.0 g 64.0 g of = -------- CH OH, were % % Yield methanol, 100 = 87.5 expected to be 64.0 g produced in the reaction 2 3 Example: Benzene, C H , reacts with bromine to form bromobenzene, C H Br, and hydrogen bromide. Reaction of 8.00 g benzene with= 12.85 g C6H5Bryielded 12.85 actual yield excess bromine g bromobenzene. Calculate % yield of bromobenzene. C6H6 + Br2 C6H5Br + HBr To calculate %yield we need to know the theoretical yield. We will use 8.0 g C6H6 (LR) because bromine is the excess reagent C H + Br C H Br + HBr MW (C6H6) = 6 x 12.011 + 6 x 1.008 = 78.11 g/mol ( How many moles of C6H6? 1mol C6H6 8.00 g C6H6 x ------------ = 0.102 mol C6H6 78. 11 g C6H6 How many moles of C6H5Br are produced? 1mol C6H5Br 0.102 mol C6H6 x ------------ = 0.102 mol C6H5Br 1 mol C6H6 Converting moles of C6H5Br to grams... The theoretical 157.0 g C6H5Br 0.102 mol C6H5Br x -------------- = 16.0 g C6H5Br yield the percent yield of C H Br actual yield % yield = ---------------- x 100 theoretical yield 12.85 g C6H5Br % yield = -------------- x 100 = 80.3% 16.0 g C6H5Br Analysis of mineral sample The mineral cerussite is mostly PbCO3, but other substances are present. To analyze for the PbCO3 content, a sample of mineral is first treated with nitric acid to dissolve PbCO3 PbCO3(s) + 2 HNO3(aq) Pb(NO3)2(aq) + H2O(l) + CO2(g) On adding sulfuric acid, lead(II) sulfate precipitates. Pb(NO3)2(aq) + H2SO4(aq) PbSO4(s) + 2 HNO3(aq) Solid PbSO4(s) is isolated and weighed. Suppose a Calculating moles and grams First we calculate moles of PbSO4 (FW = 303.3) 1 mol PbSO4 0.628 g PbSO4x ------------ = 0.00207 mol PbSO4 303.3 g PbSO4 Now, we determine moles of PbCO3 1 mol Pb(NO3)2 1 mol PbCO3 0.00207 mol PbSO4x-------------- x -------------- 1 mol PbSO4 1 mol Pb(NO3)2 = 0.00207 mol PbCO3 that must be converted to g grams and % of PbCO (FW = 267.2 g/mol) g PbCO3 out of mol PbCO3 0.00207 mol PbCO3 x ------------ = 0.553 g PbCO3 1 mol PbCO3 Now the % PbCO3 in the mineral sample (0.583 g) 0.553 g PbCO3 Mass percent PbCO3= -------------- x 100 = 94.9% 0.583 g sample 267.2 g PbCO3 Combustion Analysis sample is burned in oxygen C CO2 (g) H H2O(g) Example: compound containing only C, H, & O combustion of 28.64 mg sample of compound 88.02 mg CO2 (produced after combustion) 27.03 mg H2O (produced after combustion) MW = 286.5 (given info) determine molecular formula Example: C xH y O z + O 2 CO2 + H2 O C CO2, H H2O, O CO2 & H2O det. mass of C and of H det. mass of O (C + H + O = 100%) calc. mol C, H, & O det. empirical formula det. molecular formula Example: 1 mmol CO2 1 mmol C 88.02 mg CO2 x ------------ x ------------ x 44.01 mg CO2 1 mmol CO2 12.01 mg C x ------------ = 24.02 mg C 1 mmol C ... now H Example: 1 mmol H2O 2 mmol H 27.03 mg H2O x ------------ x ------------ x 18.02 mg H2O 1 mmol H2O 1.008 mg H x ------------ = 3.02 mg H 1 mmol H ... now mg of oxygen mg sample = mg C + mg H + mg O 28.64 mg = 24.02 mg C + 3.024 mg H + mg O mg O = 28.64 24.02 3.024 = 1.60 mg O Example: 1 mmol C 24.02 mg C x ---------- = 2.000 mmol C 12.01 mg 1 mmol H 3.024 mg H x ---------- = 3.000 mmol H 1.008 mg H 1 mmol O 1.60 mg O x ---------- = 0.100 mmol O 16.00 mg O Now, we will divide all by the smallest... Example: C2.0H3.0O0.1 0.1 0.1 0.1 C H O is the empirical formula weight of E.F. = 20 12.01 + 30 1.008 + 16.00 = 286.4 n = ------------ = ---------- = 1 MW 286.5 ...
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This note was uploaded on 08/24/2008 for the course CHM 2045 taught by Professor Geiger during the Summer '08 term at University of Central Florida.

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