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Stoichiometry Review and Practice Exercises

A comprehensive review of stoichiometry concepts with practice exercises and solutions provided. Topics include balancing chemical equations and calculating theoretical yields. Enhance your understanding with these guided exercises.

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Stoichiometry Review and Practice Exercises

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  1. Stoichiometry Chapter 9 Review

  2. Pages 73 – 74 • B • D • A • C

  3. Pages 73 – 74 • (see below) • 2 N2O + 3 O2 4 NO2 • Skip • 15 mol O2 • True • false

  4. Pages 73 – 74 • (see below) • Skip

  5. Pages 73 – 74 • 4 NH3 + 6 NO  5 N2 + 6 H2O • Skip • Skip 0.240 NH3 + 0.360 NO  0.300 N2 + 0.360 H2O 0.360 mol NO SKIP 8

  6. Pages 75 – 76 • 2 KClO3 + heat  2 KCl + 3 O2 3.0 KClO3 + heat  3.0 KCl + 4.5 O2 4.5 mol O2

  7. Pages 75 – 76 • H2 + F2 2 HF 5 H2 + 5 F2 10 HF

  8. Pages 75 – 76 • 2 H2O  2 H2 + O2 0.033 H2O  0.033 H2 + 0.0165 O2

  9. Pages 75 – 76 • 2 Na + 2 H2O  2 NaOH + H2 0.869 Na + 0.869 H2O  0.869 NaOH + 0.434 H2

  10. Pages 75 – 76 • (a) LiClO4 LiCl + 2 O2 0.941 LiClO4 0.941 LiCl + 1.88 O2

  11. Pages 75 – 76 • (b) 0.941 LiClO4 0.941 LiCl + 1.88 O2

  12. Pages 75 – 76 • 2 NaN3 2 Na + 3 N2

  13. Pages 75 – 76 • 2 NaN3 2 Na + 3 N2

  14. Pages 75 – 76 • 2 NaN3 2 Na + 3 N2 1.934 NaN3 1.934 Na + 2.9 N2

  15. Pages 77 - 78 – # 1

  16. Pages 77 - 78 • N2 + 3 H2 2 NH3 4.0 N2 + 12.0 H2 8.0 NH3 6.0 mol N2 – 4.0 mol N2 = 2.0 mol N2 extra

  17. Pages 77 - 78 • 4.0 N2 + 12.0 H2 8.0 NH3

  18. Pages 77 - 78 • Ca(OH)2 + 2 HCl CaCl2 + 2 H2O 0.050 Ca(OH)2 + 0.10 HCl 0.050 CaCl2 + 0.10 H2O 0.10 mol HCl b. HCl

  19. Pages 77 - 78 • Ca(OH)2 + 2 HCl CaCl2 + 2 H2O 0.040 Ca(OH)2 + 0.080 HCl 0.040 CaCl2 + 0.080 H2O

  20. Pages 77 - 78 • N2 + 2 O2 2 NO2 3 NO2 + H2O 2 HNO3 + NO 15 N2 + 30 O2 30 NO2

  21. Pages 77 - 78 • N2 + 2 O2 2 NO2 3 NO2 + H2O 2 HNO3 + NO 15 N2 + 30 O2 30 NO2 30 NO2 + 10 H2O 20 HNO3 + 10 NO

  22. Pages 77 - 78 • 30 NO2 + 10 H2O 20 HNO3 + 10 NO

  23. Pages 77 - 78 • 15 N2 + 30 O2 30 NO2

  24. Pages 79 – 80 • C3H4 + x O2 3 CO2 + 2 H2O • 4 • 40.0 g C3H4/mol • skip • 0.20 mole C3H4 • 3z

  25. Pages 79 – 80 • (see below) • All of the limiting reactant is turned into products with no losses. • Determine theoretical yield • Smaller

  26. Pages 79 – 80 • (see below) • 4 mol NH3 • 8.5 g NH3 • You must have more than 0.3 moles H2 • 75%

  27. Pages 79 – 80 • (see below) • 0.0692 mol HgO • 0.0346 mol O2 • 1.11 g O2 • 0.787 L O2 • 1.05 g O2

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