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Chapter3_key - Chem127 DrAbel 1 Chapter 3 Worksheet Suggested Chapter 3 book problems 2 8 10 14 18 23 25 27 29 31 33 35 37 39 41 44 46 50 51 53 63

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Unformatted text preview: Chem127 DrAbel 1 Chapter 3 Worksheet Suggested Chapter 3 book problems: 2, 8, 10, 14, 18, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 44, 46, 50, 51, 53, and 55, 61, 63, 67, 71, 73, 75, 77, 81, 83, 89, 92, 94, 96, 98, 100, 102, and 120 1. The controversial artificial sweetener saccharin has the molecular formula C3H5O3NS. What is the mass in grams of one molecule? 135.1gC2 H 5O3 NS 1mole " = 2.243x10#22 g /molecule mole 6.022x10 23 molecules 2. An Alka Seltzer tablet contains 324 mg of aspirin (C9H8O4), 1.9 g of NaHCO3 and 1.0 g of citric acid (C6H8O7). Calculate the number of molecules of aspirin in the tablet. ! 0.324gC9 H 8O4 1mole 6.022x10 23 molecules " = 1.08x10 21 molecules 180.09gC9 H 8O4 mole 3. How many moles of nitrogen atoms are in 51.7 g of ammonium nitrate? ! 51.7gNH 4 NO3 1moleNH 4 NO3 2molesN " " = 1.29moles of N atoms mole 80.02g 1moleNH 4 NO3 ! 4. What are the total number of moles of atoms in 62.4 g of silver (I) sulfate? moleAg2 SO4 7 moles of atoms 62.4gAg2 SO4 "" " = 1.40 moles 311.8gAg2 SO4 1 mole Ag2 SO4 5. Sodium perborate, NaBO3, is an "oxygen bleach" that releases oxygen, which has the bleaching action. How many grams of NaBO3 are in 4.65 moles? 4.65molesNaBO3 " 81.8gNaBO3 = 3.80x10 2 gNaBO3 1moleNaBO3 ! ! 6. The Statue of Liberty is made of 2.0 x 105 lbs of copper sheets bolted to a framework. (1 lb = 454 g) How many moles off copper are on the statue? 454g 1moleCu 2.0x10 5 lbs " " = 1.4 x10 6 molesCu 1lb 63.54gCu ! 7. Phenobarbital is C12H12N2O3. What is the molecular weight of phenobarbital? Chem127 DrAbel 2 C :12x12.01g /mol = 144.1g /mol H :12x1.008g /mol = 12.10g /mol N : 2x14.01g /mol = 28.02g /mol O : 3x16.00g /mol = 48.00g /mol 144.1+ 12.10 + 28.02 + 48.00 = 232.2g /mol 8. An insecticide has the weight percents C = 55.6%, H = 4.38%, Cl = 30.8% and O = 9.26%. The approximate molecular weight is 345. What is the molecular formula? Assume a 100g sample. Step 1, determine the # moles moleC 55.6gC " = 4.63molesC 12.01gC moleH 4.38gH " = 4.34molesH 1.01gH moleCl 30.8gCl " = 0.869molesCl 35.45gCl moleO 9.26gO " = 0.579molesO 16.00gO Step 2 : Divide By the lowest # moles (0.579) 4.63molesC /.579molesO # 8 4.34molesH /.579molesO = 7.5 0.869molesCl /.579molesO = 1.50 0.579molesO /.579molesO = 1 Step 3 : Multiply to get whole numbers 8x2 = 16 7.5x2 = 15 1.50x2 = 3 1.00x2 = 2 Empirical Formula : C16H15Cl3O 2 Step 4 : Determine the molar mass of the empirical formula Molar mass is 346g/mole, so the empirical and molecular formulas are the same 9. Balance the following equations ____ SnS2 + _6_ HCl ____ H2SnCl6 + _2_ H2S ____ Mg3B2 + __6_ H2O _3__ Mg(OH)2 + ____ B2H6 ____ (NH4)2Cr2O7 ____ Cr2O3 + _4_ H2O + ____ N2 10. Hydrogen fluoride is produced by the following reaction. How many grams of HF can be made by the reaction of 12.8 g of CaF2 and 13.2 g of H2SO4? CaF2 (s)+ H2SO4 (aq) 2 HF (aq)+ CaSO4 (aq) ! ! Chem127 DrAbel 3 First you need to determine the limiting reactant moleCaF2 2molesHF 12.8gCaF2 " " = 0.328molesHF 78.08g 1moleCaF2 moleH 2 SO4 2molesHF 13.2gH 2 SO4 " " = 0.269molesHF 98.08g 1H 2 SO4 H 2 SO4 is the limiting reagent 20.01gHF 0.269molesHF " = 5.38gHF 1moleHF ! 11. Hydrogen chloride can be prepared in the laboratory by the following reaction. If 10.0 g of NaCl and 10.0 g of H2SO4 are used, how many g of HCl can be produced? NaCl (s) + H2SO4 (aq) NaHSO4 (aq) + HCl (aq) You can do this problem using the same steps in problem 11, or below is an alternative way to determine the limiting reactant: First you need to determine the limiting reagent (or reactant), so you need to determine the number of moles of each reactant mole 10.0gNaCl " = 0.171moles 58.44g mole 10.0gH 2 SO4 " = 0.102moles 98.08g Using the stoichiometric ratios from balanced equation : 1moleH 2 SO4 0.171molesNaCl " = 0.171molesH 2 SO4 needed, 1molesNaCl we only have 0.102 moles, so H 2 SO4 is the limiting reagent 1molesHCl 36.46gHF 0.102molesH 2 SO4 " " = 3.72gHCl 1molesH 2 SO4 1moleHCl Nitroglycerin (C3H5N3O9) is a powerful explosive. How many grams of CO2 are produced from 10.0 g of nitroglycerin? (reaction not balanced!) 4C3H5N3O9 (g) 6N2 (g) + 12CO2 (g) + 10H2O (g) + O2 (g) 1mole nitroglycerin 12molesCO2 44.01g 10.0g nitroglycerin " " " = 5.81gCO2 227.11g nitroglycerin 4moles nitroglycerin 1moleCO2 12. 13. Calculate the molarity of a solution of 91 g of magnesium chloride in 2.0 L of solution. ! ! 91gMgCl2 " mole = 0.976moles 93.21g 0.976moles = 0.49M 2.0L ! 14. How many moles of lithium chlorate are in 153 mL of a 1.764 M solution? 1.764moles " .153L = 0.270moles L ! Chem127 DrAbel 4 15. Determine the mass (g) of solute required to form 250.0 mL of a 0.250 M sodium cyanide. 0.250molesNaCN 49.01g " .2500L " = 3.06gNaCN L 1moleNaCN 16. A 25.0 mL sample of concentrated HCl (12.0 M) is diluted to a final volume of 750.0 mL. What is the molarity of the final solution? M iVi = M f V f (12.0M)(25.0mL) = M f (750.0mL) M f = 0.400M ! ! 17. A solution is prepared by dissolving 516.5 mg of oxalic acid (C2H2O4) to make 100.0 mL of solution. A 10.00 mL portion is then diluted to 250.0 mL. What is the molarity of the final solution? 516.5mgC2 H 2O4 " M= 1g 1mole " = 5.74 x10#3 moles 1000mg 90.04g 5.74 x10#3 moles = 5.74 x10#2 M 0.1000L M iVi = M f V f (5.74 x10#2 M)(10.00mL) = M f (250.0mL) M f = 2.29x10#3 M ! 18. What volume (mL) of concentrated HI (5.51 M) should be used to prepare 125.0 mL of a 0.100 M solution? M iVi = M f V f (5.51M)Vi = (.100M)(125.0mL) Vi = 2.27mL 19. Describe how to make 100.0 mL of a 0.125M solution of potassium chromate from a 0.872 M potassium chromate. M iVi = M f V f ! (0.872M)Vi = (.125M)(100.0mL) Vi = 14.3mL Add 14.3mL KCr2O 7 to 85.7mL of water to make a 100.0mL solution ! ...
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This note was uploaded on 06/08/2008 for the course CHEM 127 taught by Professor Able during the Spring '08 term at Cal Poly.

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