University:
Cornell University
Course:
CHEM 1729 | Solve It!
Academic year:

2023
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198

Pages:

12
Author:

discovermsmbio

Name κεν Hour Stoich Practice A correctly balanced equation and all work must be shown! 1. Copper (II) oxide decomposes into copper and oxygen gas. What mass of copper will be produced by the decomposition of 120.5 g CuO? 2 CuO2Cu + Oz 120.5g Cuo Imal Cul 2mol Cu/63.55gcu 79.55 gruo 2mol Cuo Imol Cu 1-96.26 glu 2. Ammonia (NH3) is produced by the reaction of nitrogen gas and hydrogen gas. How much ammonia will be produced if 22.0 g of hydrogen gas react with excess nitrogen gas? N2+3Hz→ 2NH3 22.0 GHz Imol Hz 1202ghz 2 mol NH3 3mol H2 17.04 ANH 3 Imol 3 = 123.72 g NH3 3. The reaction of sodium and water produces sodium hydroxide and hydrogen gas. What mass of hydrogen gas is produced if 17.54 g of NaOH is produced by the reaction? 2 Na + 21120 → 2 Na OH H2 17.SAg NaOH Imol NaOH 40 g NaOH Imol H2 2mol NaOlt i mal fiz 2 =0.44 ghz/ 2 2,02 4. The combustion of acetic acid (HC₂H₃O₂) produces carbon dioxide and water. What mass of carbon dioxide will be produced from the combustion of 25.0 g of acetic acid? HC2H302 +20 2 →2002 + 2H2O 25g HC2H302 Imol HCzH302 2 mol CO2 44 0g(02 36.64 g CO2/ 5. 20.0 g of iron (III) sulfide was prepared by heating iron and excess sulfur. What mass of iron was used in the preparation? 2 Fe + 35 Fe253 20.0g Fez Sz Imol Fez53 2mol Fe 1m61 Fez Sz Ss.85 gre Imol Fe ۱۰ 207.91gFez Sz 10.75g Fe b) Limiting Reactant and Stoichiometry 13.6 g NH₃ (1 mol NH₃ / 17.04 g NH₃) * (1 mol P₂O₅ / 2 mol NH₃) = 0.4 mol P₂O₅ 30.8 g CO₂ (1 mol CO₂ / 44.01 g CO₂) * (1 mol P₂O₅ / 2 mol CO₂) = 0.35 mol P₂O₅ Limiting Reactant and Set Up: NH₃ + CO₂ → (NH₂)₂CO + H₂O a) 13.6 g NH₃ (1 mol NH₃ / 17.04 g NH₃) * (1 mol urea / 2 mol NH₃) = 0.4 mol urea b) 30.8 g CO₂ (1 mol CO₂ / 44.01 g CO₂) * (1 mol urea / 1 mol CO₂) = 0.7 mol urea 2. 3H₂ + N₂ → 2NH₃ a) 3.8 g H₂ (1 mol H₂ / 2.02 g H₂) * (2 mol NH₃ / 3 mol H₂) = 0.99 mol NH₃ b) 28 g N₂ (1 mol N₂ / 28.02 g N₂) * (2 mol NH₃ / 1 mol N₂) = 2 mol NH₃ b) 0.99 mol NH₃ * (17.04 g NH₃ / 1 mol NH₃) = 16.87 g NH₃ 2 mol - 0.99 mol = 1.01 mol N₂ * (28.02 g N₂ / 1 mol N₂) = 14.15 g N₂ 3. a) 2CrPO₄ + 3H₂ → 2Cr + 2H₃PO₄ 29.4 g CrPO₄ (1 mol CrPO₄ / 146.97 g CrPO₄) * (2 mol Cr / 2 mol CrPO₄) = 0.2 mol Cr 2 g H₂ (1 mol H₂ / 2.02 g H₂) * (2 mol Cr / 3 mol H₂) = 0.33 mol Cr b) 1.66 mol - 0.2 mol = 1.46 mol Cr * (52 g Cr / 1 mol Cr) = 75.92 g Cr 2.02 g H₂ (1 mol H₂ / 2.02 g H₂) * (2 mol H₂O / 3 mol H₂) = 1.34 g H₂O 4. a) 2CaTiO₃ + 13O₂ → 8CO₂ + 10H₂O 11 g CaTiO₃ (1 mol CaTiO₃ / 158.14 g CaTiO₃) * (8 mol CO₂ / 2 mol CaTiO₃) = 0.76 mol CO₂ 4.6 g O₂ (1 mol O₂ / 32.00 g O₂) * (8 mol CO₂ / 13 mol O₂) = 0.92 mol CO₂ b) 0.76 mol CO₂ * (44.01 g CO₂ / 1 mol CO₂) = 33.45 g CO₂ 0.76 mol CO₂ * (1 mol H₂O / 8 mol CO₂) * (18.02 g H₂O / 1 mol H₂O) = 17.12 g H₂O c) 0.92 mol - 0.76 mol = 0.16 mol CO₂ * (32 g O₂ / 1 mol CO₂) = 8.32 g O₂ Chemistry Practice 1. 3 F₂ + 2 Al → 2 AlF₃ 100 g AlF₃ * (1 mol AlF₃ / 83.98 g AlF₃) * (3 mol F₂ / 2 mol AlF₃) * (37.98 g F₂ / 1 mol F₂) = 167.86 g F₂ 2. (NH₄)₂Cr₂O₇ → Cr₂O₃ + 4H₂O + N₂ 25 g (NH₄)₂Cr₂O₇ * (1 mol (NH₄)₂Cr₂O₇ / 252.07 g (NH₄)₂Cr₂O₇) * (1 mol Cr₂O₃ / 1 mol (NH₄)₂Cr₂O₇) * (152.01 g Cr₂O₃ / 1 mol Cr₂O₃) = 15.07 g Cr₂O₃ 3. H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O 15 g H₂SO₄ * (1 mol H₂SO₄ / 98.09 g H₂SO₄) * (2 mol NaOH / 1 mol H₂SO₄) * (40.00 g NaOH / 1 mol NaOH) = 12.23 g NaOH 4. C₆H₆ + 15/2 O₂ → 12CO₂ + 3H₂O 156.24 g C₆H₆ * (1 mol C₆H₆ / 78.11 g C₆H₆) * (12 mol CO₂ / 1 mol C₆H₆) * (6.022 × 10²³ molecules CO₂ / 1 mol CO₂) = 7.22 × 10²⁴ molecules CO₂ 5. Al₂(SO₄)₃ + 3BaCl₂ → 3BaSO₄ + 2AlCl₃ 25 g BaSO₄ * (1 mol BaSO₄ / 233.39 g BaSO₄) * (3 mol BaCl₂ / 3 mol BaSO₄) * (208.23 g BaCl₂ / 1 mol BaCl₂) = 22.30 g BaCl₂

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