Mole Calculation Worksheet - Answer Key PDF

 Mole Calculation Worksheet – Answer Key 1) How many moles are in 15 grams of lithium? 2.2 moles 2) How many grams are in 2.4 moles of sulfur? 77 grams if S(s), 6.2 x 102 grams if S8(s) 3) How many moles are in 22 grams of argon? 0.55 moles 4)...

 Mole Calculation Worksheet – Answer Key 1) How many moles are in 15 grams of lithium? 2.2 moles 2) How many grams are in 2.4 moles of sulfur? 77 grams if S(s), 6.2 x 102 grams if S8(s) 3) How many moles are in 22 grams of argon? 0.55 moles 4) How many grams are in 88.1 moles of magnesium? 2.14 x 103 grams 5) How many moles are in 2.3 grams of phosphorus? 0.074 moles (0.019 if P4(s)) 6) How many grams are in 11.9 moles of chromium? 619 grams 7) How many moles are in 9.8 grams of calcium? 0.24 moles 8) How many grams are in 238 moles of arsenic? 1.78 x 104 grams What are the molecular weights of the following compounds? 9) NaOH 40.00 g/mol 12) H3PO4 98.00 g/mol 10) H2O 18.02 g/mol 13) Mn2Se7 662.60 g/mol 11) MgCl2 95.21 g/mol 14) (NH4)2SO4 132.17 g/mol 15) How many grams are in 4.5 moles of sodium fluoride, NaF? 1.9x102 grams 16) How many moles are in 98.3 grams of aluminum hydroxide, Al(OH)3? 1.26 moles 17) How many grams are in 0.02 moles of beryllium iodide, BeI2? 5 grams 18) How many moles are in 68 grams of copper (II) hydroxide, Cu(OH)2? 0.70 moles 19) How many grams are in 3.3 moles of potassium sulfide, K2S? 3.6x102 grams 20) How many moles are in 1.2 x 103 grams of ammonia, NH3? 70 moles 21) How many grams are in 2.3 x 10-4 moles of calcium phosphate, Ca3(PO3)2? 0.064 grams 22) How many moles are in 3.4 x 10-7 grams of silicon dioxide, SiO2? 5.7 x 10-9 moles 23) How many grams are in 1.11 moles of manganese sulfate, Mn3(SO4)7? 929 grams © 2000 Cavalcade Publishing – All Rights Reserved For chemistry help, visit http://www.chemfiesta.com  Molar Ratio Practice Problems Solutions Following each equation are two requests for molar ratios from the equation. 1) N2 + 3 H2 → 2 NH3 N2 to H2: 1:3 NH3 to H2: 2:3 2) 2 SO2 + O2 → 2 SO3 O2 to SO3: 1:2 O2 to SO2: 1:2 3) PCl3 + Cl2 → PCl5 PCl3 to Cl2: 1:1 PCl3 to PCl5: 1:1 4) 4 NH3 + 3 O2 → 2 N2 + 6 H2O NH3 to N2: 4:2 H2O to O2: 6:3 5) Fe2O3 + 3 CO → 2 Fe + 3 CO2 CO to CO2: 3:3 Fe to CO: 2:3 Mole to Mole Practice Problems Here's the equation to use for all three problems: 2 H2 + O2 → 2 H2O 1) How many moles of H2O are produced when 5 moles of oxygen are used? Remember: Starting with five moles of oxygen and based on the balanced equation, for every 1 mole of oxygen used, two moles of water are produced. 5 _______ mol O2 x 2 mol H2O = 10 mol H2O 1 mol O2 water 2) If 3 moles of H2O are produced, how many moles of oxygen must be consumed? Answer: 1.5 mol O2 Sig Digs: 2 mol O2 3) If 2.5 moles of H2O are produced, how many moles of hydrogen gas must be used? Answer: 2.5 mol H2 4) Suppose 4.0 grams of H2 were used? How many grams of water would be produced? Answer: 36.0 grams H2O Sig Digs: 36g H2O  CHAPTER 7 BLM 7.2.3A ANSWER KEY Gravimetric Stoichiometry Problems Answer Key 1. 2Fe(s) + 3Cl2(g) → 2FeCl3(s) 2 mol Fe (a) nFe = 1.3mol Fe × = 0.87 mol Fe(s) 3mol Cl2 3mol Cl2 g (b) mCl2 = 4.65 mol Fe × × 70.90 Cl2 = 495g Cl2 (g) 2 mol Fe mol mol 2 mol FeCl3 g (c) mFeCl3 = 0.520 g Cl2 × Cl2 × × 162.20 FeCl3 = 0.793g FeCl3 (s) 70.90 g 3mol Cl2 mol 2. (a) 2C2H6(g) + 7O2(g) → 4CO2(g) + 6 H2O(g) 7 mol O 2 (b) nO2 = 10.8 mol C 2 H 6 × = 37.8 mol O 2 (g) 2 mol C 2 H 6 4 mol CO 2 g (c) mCO2 = 0.550 mol O 2 × × 44.01 CO 2 = 13.8g CO 2 (g) 7 mol O 2 mol mol 2 mol C2 H 6 g (d) mC2 H6 = 10.6 g H 2 O × H 2O × × 30.08 C2 H 6 = 5.90 g C 2 H 6 (g) 18.02 g 6 mol H 2 O mol 3. (a) Xe(g) + 2F2(g) → XeF4(g) 1mol Xe (b) nXe = 3.54 × 10−1 mol F2 × = 0.177 mol Xe(g) 2 mol F2 mol 1mol XeF4 g (c) mXeF4 = 4.35g F2 × F2 × × 207.29 XeF4 = 11.9 g XeF4 (g) 38.00 g 2 mol F2 mol mol 1mol Xe g (d) mXe = 15.7 g F2 × F2 × × 131.29 Xe = 27.1 g Xe(g) 38.00 g 2 mol F2 mol Copyright © 2007, McGraw-Hill Ryerson Limited, a Subsidiary of the McGraw-Hill Companies. All rights reserved. This page may be reproduced for classroom use by the purchaser of this book without the written permission of the publisher. CHAPTER 8 BLM 8.1.2A ANSWER KEY Limiting Reactant Problems Answer Key 1. ZnO(s) + C(s) → Zn(s) + CO(g) 1mol Zn 17.2 mol ZnO × = 17.2 mol Zn(s) 1mol ZnO 1mol Zn 43.2 mol C × = 43.2 mol Zn(s) 1mol C Zinc oxide is the limiting reactant. 2. C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(g) 6 mol CO 2 5.55 mol C6 H12 O6 × = 33.3mol CO 2 (g) 1mol C6 H12 O6 6 mol CO 2 34.0 mol O 2 × = 34.0 mol CO 2 (g) 6 mol O 2 Glucose is the limiting reactant. This seems reasonable because oxygen is quite often in abundance. 3. 4C3H6(g) + 6NO(g) → 4C3H3N(g) + 6H2O(g) + N2(g) mol 4 mol C3 H 3 N 126 g C3H 6 × × = 2.99 mol C3H 3 N(g) 42.09 g 4 mol C3 H 6 mol 4 mol C3 H 3 N 175g NO × × = 3.89 mol NO(g) 30.01g 6 mol NO(g) C3H6 is the limiting reactant. 4. CaF2(s) + H2SO4(aq) → CaSO4(s) + 2HF(g) mol 1mol CaSO 4 10.0 g CaF2 × × = 0.128 mol CaSO 4 (s) 78.08g 1mol CaF2 mol 1mol CaSO 4 15.5g H 2SO 4 × H 2SO 4 × = 0.158 mol CaSO 4 (s) 98.09 g 1mol H 2SO 4 Calcium fluoride is the limiting reactant. 5. Li3N(s) + 3H2O(ℓ) → 3LiOH(aq) + NH3(g) mol 1mol NH 3 4.87 g Li3 N × Li3 N × = 0.140 mol NH 3 (g) 34.83g 1mol Li3 N mol 1mol NH 3 7.74 g H 2 O × H 2O × = 0.143mol NH 3 (g) 18.02 g 3mol H 2 O Lithium nitride is the limiting reactant. Copyright © 2007, McGraw-Hill Ryerson Limited, a Subsidiary of the McGraw-Hill Companies. All rights reserved. This page may be reproduced for classroom use by the purchaser of this book without the written permission of the publisher. CHAPTER 8 BLM 8.1.2A ANSWER KEY Limiting Reactant Problems Answer Key (continued) 6. 2Al(s) + 3CuCl2(aq) → 3Cu(s) + 2AlCl3(aq) mol 2 mol AlCl3 0.25g Al × Al × = 9.3 × 10−3 mol AlCl3 (aq) 26.98g 2 mol Al mol 2 mol AlCl3 0.51g CuCl2 × CuCl2 × = 2.5 × 10−3 mol AlCl3 (aq) 134.45g 3mol CuCl2 Copper(II) chloride is limiting. 7. Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) mol 1mol ZnCl2 33.76 g Zn × Zn × = 0.5161mol ZnCl2 (aq) 65.41g 1mol Zn mol 1mol ZnCl2 54.08g HCl × HCl × = 0.7416 mol ZnCl2 (aq) 36.46 g 2 mol HCl Hydrogen chloride is in excess. (Zinc is limiting.) 8. ClO2(g) + H2O(ℓ) → HCl(aq) + HClO3(aq) mol 1mol HCl 71.00 g ClO 2 × ClO 2 × = 1.053mol HCl(aq) 67.45g 1mol ClO 2 mol 1mol HCl 19.00 g H 2 O × H 2O × = 1.054 mol HCl(aq) 18.02 g 1mol H 2 O Chloride dioxide is limiting. Copyright © 2007, McGraw-Hill Ryerson Limited, a Subsidiary of the McGraw-Hill Companies. All rights reserved. This page may be reproduced for classroom use by the purchaser of this book without the written permission of the publisher. CHAPTER 8 BLM 8.1.3A ANSWER KEY Expected Quantity of Product Problems Answer Key 1. 2MnI2(s) + 13F2(g) → 2MnF3(s) + 4IF5(ℓ) mol 2 mol MnF3 g 1.24 g MnI 2 × MnI 2 × × 111.94 MnF3 = 0.450 g MnF3 (s) 308.74 g 2 mol MnI 2 mol mol 2 mol MnF3 g 25.0 g F2 × F2 × × 111.94 MnF3 = 11.3g MnF3 (s) 38.00 g 13mol F2 mol MnI2(s) is the limiting reactant, and the expected mass of manganese(III) fluoride is 0.450g. 2. 3Cu(s) + 8HNO3(aq) → 3Cu(NO3)2(aq) + 2NO(g) + 4H2O(ℓ) mol 2 mol NO g 57.4 g Cu × Cu × × 30.01 NO = 18.1g NO(g) 63.55g 3mol Cu mol mol 2 mol NO g 165g HNO3 × HNO3 × × 30.01 NO = 19.6 g NO(g) 63.02 g 8 mol HNO3 mol Copper is the limiting reactant, and the expected mass of NO(g) is 18.1g. 3. 2H2O2(ℓ) → 2H2O(ℓ) + O2(g) mol 2 mol H 2 O g 10.0 g H 2 O 2 × H 2O 2 × ×18.02 H 2 O = 5.30 g H 2 O (ℓ) 34.02 g 2 mol H 2 O 2 mol The expected mass of water is 5.30 g. 4. 4NH3(g) + 7O2(g) → 4NO2(g) + 6H2O(g) mol 4 mol NO 2 g 1.30 ×103 g NH 3 × × × 46.01 NO 2 = 3.51× 103 g = 3.51kg NO 2 (g) 17.04 g 4 mol NH 3 mol mol 4 mol NO 2 g 4.21×103 g O 2 × O2 × × 46.01 NO 2 = 3.46 × 103 g = 3.46 kg NO 2 (g) 32.00 g 7 mol O 2 mol The expected mass of nitrogen dioxide is 3.46 kg. Oxygen gas is limiting. 5. Fe2(SO4)3(s) + 6NaOH(aq) → 2Fe(OH)3 (aq)+ 3Na2SO4(aq) mol 2 mol Fe(OH)3 g 10.0 g Fe 2 (SO 4 )3 × Fe 2 (SO 4 )3 × ×106.88 Fe(OH)3 = 5.35g Fe(OH)3 (aq) 399.91g 1mol Fe 2 (SO 4 )3 mol mol 2 mol Fe(OH)3 g 10.0 g NaOH × NaOH × × 106.88 Fe(OH)3 = 8.91g Fe(OH)3 (aq) 40.0 g 6 mol NaOH mol The expected mass of iron(III) hydroxide is 5.35g. Iron(III) sulfate is limiting. Copyright © 2007, McGraw-Hill Ryerson Limited, a Subsidiary of the McGraw-Hill Companies. All rights reserved. This page may be reproduced for classroom use by the purchaser of this book without the written permission of the publisher. CHAPTER 8 BLM 8.2.1A ANSWER KEY Percentage Yield Problems Answer Key Experimental yield 1. Percentage yield = × 100% Predicted yield 1.72 g % yield = × 100% 9.10 g % yield = 18.9% 2. HBrO3(aq) + 5HBr(aq) → 3H2O(ℓ) + 3Br2(aq) mol 3mol Br2 g (a) 20.0 g HBrO3 × × × 159.80 Br2 = 74.4 g Br2 128.91g 1mol HBrO3 mol The predicted yield of Br2 is 74.4 g. (b) Percentage yield = Experimental yield × 100% Predicted yield 47.3g % yield = × 100% 74.4 g % yield = 63.6% 3. PbCl2(s) + Na2CrO4(aq) → PbCrO4(s) + 2NaCl(aq) mol 1mol PbCrO 4 g (a) 12.5 g PbCl 2 × × × 323.20 PbCrO 4 = 14.5g PbCrO 4 278.10 g 1mol PbCl2 mol The predicted yield of lead(II) chromate is 14.5 g. Experimental yield (b) Percentage yield = × 100% Predicted yield 13.8g % yield = ×100% 14.5g % yield = 95.2% Copyright © 2007, McGraw-Hill Ryerson Limited, a Subsidiary of the McGraw-Hill Companies. All rights reserved. This page may be reproduced for classroom use by the purchaser of this book without the written permission of the publisher. CHAPTER 8 BLM 8.2.1A ANSWER KEY Percentage Yield Problems Answer Key (continued) 4. CaCO3(s) + 2HCl(aq) → CaCl2(s) + CO2(g) + H2O(ℓ) mol 1mol CO 2 g 15.7 g CaCO3 × × × 44.01 CO 2 = 6.90 g CO 2 100.09 g 1mol CaCO3 mol The expected yield of carbon dioxide is 6.90 g. Experimental yield Percentage yield = × 100% Predicted yield Experimental yield = ( Percentage yield )( Predicted yield ) 100% Experimental yield = ( 6.90 g )(81.5% ) 100% Experimental yield = 5.62 g CO 2 (g) Copyright © 2007, McGraw-Hill Ryerson Limited, a Subsidiary of the McGraw-Hill Companies. All rights reserved. This page may be reproduced for classroom use by the purchaser of this book without the written permission of the publisher.