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Stoichiometry: Review Questions for AP Chemistry (page 2)

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By — McGraw-Hill Professional
Updated on Feb 1, 2011

Answers and Explanations

There are multiple "correct" ways to do these calculations. Only one calculation is shown for each answer.

  1. D—The reaction is H2SO4 + 2 KOH → K2SO4 + 2 H2O
  2. C—The reaction is H2C2O4 + 2 NaOH → Na2 C2O4 + 2 H2O
  3. C—Moles acid = (50.0 mL)(0.20 mol acid/1000 mL) = 0.0100 mol
  4. Moles base = (50.0 mL)(0.20 mol base/1000 mL) = 0.0100 mol

    There is sufficient base to react completely with only one of the ionizable hydrogens from the acid. This leaves H2AsO4.

  5. AOx = oxidizing agent = Cr2O7 2–; Red = reducing agent = Fe2+
  6. C—The reaction is WO3 + 3 H2→W + 3 H2O
  7. (0.0500 mol WO3)(l mol W/l mol WO3) (183.8 g W/1 mol W) = 9.19 g W

  8. D— 63.2% Mn leaves 36.8% O
  9. 63.2/54.94 = 1.15 Mn 36.8/16.0 = 2.30 O

    Thus, there is 1 Mn/2 O.

  10. C—V: 2.39/50.94 = 0.0469 O:1.00/16.0 = 0.0625
  11. 0.0469/0.0469 = 1 0.0625/0.0469 = 1.33

    Multiplying both by three gives: 3 V and 4 O.

  12. A—(25.0 g(NH4)2SO4)(l mol(NH4)2SO4/132 g) × (2 mol N/l mol(NH4)2SO4)(14.0 g N/1 mol N) = 5.30 g
  13. C—[(6 mol H2O)(18 g/mol H2O)]/(250 g Na2SO4 · 6 H2O) × 100% = 43%
  14. B—Calculate the moles of acid to compare to the moles of Cu: (10.0 mL)(12 mol/1000 mL) = 0.12 mol
  15. The acid is the limiting reactant, and will be used to calculate the moles of NO formed.

    (0.12 mol acid)(2 mol NO/8 mol acid) = 0.030 mol

  16. E—The balanced chemical equation is:
  17. 2 Au2O3 → 4 Au + 3 O2 (221 g Au2O3)(l mol Au2O3/442 g Au2O3)(3 mol O2/2 mol Au2O3) = 0.750 mol O2

  18. D—The balanced equation is:
  19. 4 C4H11N(l) + 27 O2(g) → 16 CO2(g) + 22 H2O(1) + 2 N2(g)

  20. E—The KMnO4 is the limiting reagent. Each mole of KMnO4 will produce a mole of MnSO4.
  21. D—The balanced equation is:
  22. 2 KClO3 → 2 KCl + 3 O2 (4.0 mol KClO3)(3 mol O2/2 mol KCIO3) = 6.0 mol O2

  23. E—The balanced equation is:
  24. 2 C8H18 + 25 O2 → 16 CO2 + 18 H2O (1.00 mol C8H18)(25 mol O2/2 mol C8H18) = 12.5 mol O2

  25. E—(0.200 mol Ca)(l mol H2/l mol Ca)(22.4 L at STP/l mol H2)= 4.48 L
  26. C—There are 5.00 × 10–3 mol of CrO42– and an equal number of mol of SnO22–. Thus SnO22– is the limiting reactant (larger coefficient in the balanced reaction).
  27. (5.00 × 10–3 mol SnO22–)(2 mol OH/3 mol SnO22–) = 3.33 × 10–3 mol OH

  28. D—The volume of water is irrelevant.
  29. 0.20 mol of KBr will require 0.10 mol of Pb(NO3)2

    0.20 mol of MgBr2 will require 0.20 mol of Pb(NO3)2

    Total the two yields.

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