Chapter 11 Homework Answers

Mole Ratio Practice Pb + 2HCl  PbCl2 + H2 C + 2ZnO  2Zn + CO2 x mol Pb + 2HCl  PbCl2 + H2 C + 2ZnO  2Zn + CO2 1 mol Pb = 2 mol HCl 0.38 mol x mol 2 mol ZnO = 1 mol CO2

Mole Ratio Practice 3. P4 + 6Br2  4PBr3 4. C3H8 + 5O2  3CO2 + 4H2O x mol 3. P4 + 6Br2  4PBr3 4. C3H8 + 5O2  3CO2 + 4H2O 6 mol Br2 = 4 mol PBr3 0.38 mol x mol 1 mol C3H8 = 5 mol O2

Mole Ratio Practice 5. 4Fe + 3O2  2Fe2O3 x mol 5. 4Fe + 3O2  2Fe2O3 6. 2C6H6 + 15O2  12CO2 + 6H2O 4 mol Fe = 2 mol Fe2O3 2.35 mol x mol 15 mol O2 = 6 mol H2O

Mole Ratio Practice 7. 2Al + 6HCl  2AlCl3 + 3H2 x mol 7. 2Al + 6HCl  2AlCl3 + 3H2 8. Zn + H2SO4  ZnSO4 + H2 2 mol Al = 6 mol HCl 0.36 mol x mol 1 mol Zn = 1 mol H2

Stoichiometry Practice 0.284 mol x mol 1. Mg + 2HCl  MgCl2 + H2 2. Br2 + 2NaI  2NaBr + I2 2 mol HCl = 1 mol H2 11.27 g x mol 1 mol Br2 = 2 mol NaBr

Stoichiometry Practice 2.09 mol x g 3. WO3 + 3H2  W + 3H2O 4. C3H8 + 5O2  3CO2 + 4H2O 1 mol WO3 = 3 mol H2O 6.4 L x L 5 mol O2 = 3 mol CO2

Stoichiometry Practice 3 mol CCl4= 2 mol SbCl3 x g 186g 5. 3CCl4 + 2SbF3  3CCl2F2 + 2SbCl3 6. 2Al + 6HCl  2AlCl3 + 3H2 0.46 mol x L 2 mol Al = 3 mol H2

Stoichiometry Practice 2 mol HCl = 1 mol MgCl2 x g 16.4 g 7. Mg + 2HCl  2AlCl3 + 3H2 8. 2Na + 2H2O  2NaOH + H2 x g 5.68 L 2 mol H2O = 1 mol H2

Stoichiometry Practice 227 g x mol 2 mol CO = 1 mol O2 9. 2CO + O2  2CO2 10. 2Al + 3H2SO4  Al2(SO4)3 + 3H2 2 mol H2O = 1 mol H2 x atoms 4.72 L

Mixed Review Answers 1.) 40 L NH3 2.) 2.0 g O2 3.) 2.5 x 1022 formula units KCl 4.) 0.86 L H2 5.) 0.028 mol H2O 6.) 9.03 x 1024 molecules Cl2

Stoichiometry Mixed Review 25 g x L 1 mol N2 = 2 mol NH3 1. N2 + 3H2  2NH3 2. 2KClO3  2KCl + 3O2 5.0 g x g 2 mol KClO3 = 3 mol O2

Stoichiometry Mixed Review 5.0 g x F.U. 2 mol KClO3 = 2 mol KCl 2KClO3  2KCl + 3O2 4. Zn + HCl  ZnCl2 + H2 2.5 g x L 1 mol Zn = 1 mol H2

Stoichiometry Mixed Review 2 mol H2O= 1 mol Na2SO4 x mol 2.0 g 5. H2SO4 + 2NaOH  2H2O + Na2SO4 6. 2AlCl3 2Al + 3Cl2 10 mol x molecules 2 mol AlCl3 = 3 mol Cl2

Stoichiometry Mixed Review BACK 20 g x L 7. 2K + 2H2O  2KOH + H2 8. 2Al + 6HCl  2AlCl3 + 3H2 2 mol K= 1 mol H2 30.2 g x L 2 mol Al = 2 mol HCl

Stoichiometry Mixed Review BACK 5.7 L x L 1 mol H2 = 2 mol HCl 9. H2 + Cl2  2HCl 10. 2H2 + O2  2H2O x L .45 mol 2 mol H2 = 1 mol O2

Stoichiometry Mixed Review BACK 26.9 L x L 8 mol O2 = 8 mol SO2 11. S8 + 8O2  8SO2 12. Zn + Cl2  ZnCl2 16.8 L x g 1 mol Cl2 = 1 mol ZnCl2

Limiting Reactant Practice x g 1. 6Li + N2  2Li3N Li is the limiting reactant 93.9 g Li3N produced

Limiting Reactant Practice x g 2. C3H8 + 5O2  3CO2 + 4H2O O2 is the limiting reactant 4.5 g H2O produced

Limiting Reactant Practice x g 3. KOH + HNO3  KNO3 + H2O HNO3 is the limiting reactant 19.3 g KNO3 produced

Limiting Reactant Practice x g 4. 2C3H6 + 2NH3 + 3O2  C3H3N + 6H2O O2 is the limiting reactant 20.4 g H2O produced

Percent Yield Practice 6.8 x 104 g x g 1. 2H2 + CO  CH3OH

Percent Yield Practice 30.6 g x g 2. C7H16 + 11O2  7CO2 + 8H2O

Percent Yield Practice 65.3 g x g 3. CCl4 + 2HF  CCl2F2 + 2HCl

Percent Yield Practice 58.5 g 98.0 g x g 4. 4Al + 3O2  2Al2O3

Limiting Reactant & % Yield Practice 3.45 mol 4.85 mol x mol 1. N2 + 3H2  2NH3 H2 is the limiting reactant 3.2 mol NH3 produced

Limiting Reactant & % Yield Practice 20.0 mol 10.0 mol x mol 2. 2C2H2 + 5O2  4CO2 + 2H2O O2 is the limiting reactant 8.0 mol CO2 produced

Limiting Reactant & % Yield Practice 2.36 mol 3.89 mol x g 3. CH3CO2H + NaOH  NaCH3CO2 + H2O CH3CO2H is the limiting reactant 42.5 g H2O produced

Limiting Reactant & % Yield Practice x g 4. Na2SO4 + Ba(NO3)2  2NaNO3 + BaSO4 Ba(NO3)2 is the limiting reactant 44.7 g BaSO4 produced

Limiting Reactant & % Yield Practice x g 5. 2CH3CO2H + Pb(OH)2  Pb(CH3CO2)2 + 2H2O *CH3CO2H is in excess [Pb(OH)2 is limiting] 13.5 g Pb(CH3CO2)2 produced

Limiting Reactant & % Yield Practice x g 6. 2C6H6 + 15O2  12CO2 + 6H2O

Limiting Reactant & % Yield Practice x g 7. C2H4 + Cl2  C2H4Cl2

Limiting Reactant & % Yield Practice x g 8. Al2O3 + 3H2  2Al + 3H2O

Chapter 11 Review 2KClO3  2KCl + 3O2 2Na + 2H2O  2NaOH + H2 5.8 mol x L 2KClO3  2KCl + 3O2 2Na + 2H2O  2NaOH + H2 20.0g x g

Chapter 11 Review 3. 4. H2 + F2  2HF 16.8mol x g

Chapter 11 Review 5. 2H2O  2H2 + O2 6. Ca(OH)2 + 2HCl  CaCl2 + 2H2O x g 9.2 L 5. 2H2O  2H2 + O2 6. Ca(OH)2 + 2HCl  CaCl2 + 2H2O 126g 150g x g Ca(OH)2 is the limiting reactant. 61.2 g H2O is produced.

Chapter 11 Review 73.2g 120.5g x g 7. CaCl2 + Na2CO3  CaCO3 + 2NaCl

Chapter 11 Review Limiting reactant runs out first. It determines how much product is formed. Excess reactant will have some left over when limiting runs out. 1. Convert to moles (if needed) 2. Mole:Mole Ratio 3. Convert to what the question asks for (if needed)

Chapter 11 Review Using the coefficient in the balanced chemical equation, we can relate the number of moles of each substance in the reaction. When the reactants are present in exact molar ratio. Neither is limiting or excess because they run out at the same time .

“g” and “l” don’t affect the problem! Extra Stoich Practice 1.20 mol x mol 1.) 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(l) 2.) C2H4(g) + 3 O2(g)  2 CO2(g) + 2 H2O(g) 1.25 L x L

Extra Stoich Practice X g 3.45 L 3.) CaCO3(s)  CO2(g) + CaO(s) 4.) H2SO4(aq) + 2NaOH(aq)  H2O(l) + Na2SO4(aq) x mol 2.0 g

Extra Stoich Practice 70.0 L X g 5.) 2 NaN3(s) → 2 Na (s) + 3 N2(g) 6.) 2 Cl2(g) + C2H2(g)  C2H2Cl4(l) x g 75.0 g

Extra Lim. React. & % Practice 80 g 28 g x mol 1.) I2O5(g) + 5 CO(g)  5 CO2(g) + I2(g) Part A) CO is the limiting reactant 0.20 mol I2 is produced Part B)

Extra Lim. React. & % Practice 25 g 30 g x g 2.) Zn + S  ZnS Zn is the limiting reactant 37.3 g ZnS are produced

Extra Lim. React. & % Practice 5 g 10 g x g 3.) 16Al + S8  8Al2S3 Al is the limiting reactant 13.9 g Al2S3 are produced

Extra Lim. React. & % Practice 10 g 20 g x g 4.) 3Zn + 2MoO3  Mo2O3 + 3ZnO Zn is the limiting reactant 12.4 g ZnO are produced

Extra Lim. React. & % Practice x g 25 g 5.) 3AgNO3 + FeCl3 3AgCl + Fe(NO3)3

Extra Lim. React. & % Practice x g 25 g 6.) CaCO3 + SO2 CaSO3 + CO2