- Basic Math and Science Chemical Reactions and Stoichiometry Questions and Answers Flashcards

In the reaction 2Al(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂(g), if 54 g of aluminum reacts with excess HCl, what volume of H₂ gas is produced at STP (0°C, 1 atm)?

Answer: 33.6 L

Molar mass of Al = 27 g/mol, so 54 g = 2 mol Al. The molar ratio of Al to H₂ is 2:3, so 2 mol Al produces 3 mol H₂. At STP, 1 mol of any gas = 22.4 L, so 3 mol × 22.4 L/mol = 67.2 L. Wait — let me recheck: the ratio is 2 mol Al : 3 mol H₂. 2 mol Al → 3 mol H₂. 3 mol × 22.4 L = 67.2 L. The correct answer is 67.2 L.

A reaction vessel contains 0.50 mol N₂ and 1.80 mol H₂. Using the reaction N₂ + 3H₂ → 2NH₃, which statement correctly identifies the limiting reagent and the theoretical yield of NH₃?

Answer: H₂ is the limiting reagent; yield = 1.20 mol NH₃

Check limiting reagent: 0.50 mol N₂ requires 0.50 × 3 = 1.50 mol H₂. We have 1.80 mol H₂, which is more than enough — so N₂ is NOT the limit. Now check H₂: 1.80 mol H₂ requires 1.80/3 = 0.60 mol N₂, but we only have 0.50 mol N₂. Wait — that means N₂ is actually limiting. 0.50 mol N₂ × (2 mol NH₃ / 1 mol N₂) = 1.00 mol NH₃. The correct answer is: N₂ is limiting, yield = 1.00 mol NH₃.

When 8.00 g of CH₄ is burned in 48.0 g of O₂ according to CH₄ + 2O₂ → CO₂ + 2H₂O, what is the percent yield if 19.8 g of CO₂ is actually collected? (Molar masses: C=12, H=1, O=16)

Answer: 90.0%

Moles of CH₄ = 8.00/16 = 0.500 mol. Moles of O₂ = 48.0/32 = 1.50 mol. Required O₂ for 0.500 mol CH₄ = 0.500 × 2 = 1.00 mol — we have 1.50 mol, so CH₄ is limiting. Theoretical CO₂ = 0.500 mol × 44 g/mol = 22.0 g. Percent yield = (19.8/22.0) × 100 = 90.0%.

A hydrate of copper(II) sulfate loses 36.0 g of water when 125.0 g of the hydrate is heated to drive off all water. What is the formula of the hydrate? (Molar masses: CuSO₄ = 159.6 g/mol, H₂O = 18.0 g/mol)

Answer: CuSO₄·5H₂O

Mass of anhydrous CuSO₄ = 125.0 − 36.0 = 89.0 g. Moles of CuSO₄ = 89.0/159.6 ≈ 0.5576 mol. Moles of H₂O = 36.0/18.0 = 2.00 mol. Ratio of H₂O to CuSO₄ = 2.00/0.5576 ≈ 3.59 — rounding, this does not cleanly fit 3 or 4. Let's re-examine: if ratio ≈ 3.6, closest whole number hydrate would be 4. Actually: for 5H₂O — moles CuSO₄·5H₂O: M = 159.6 + 5(18) = 249.6. Moles = 125.0/249.6 = 0.5008. Water = 5 × 0.5008 = 2.504 mol × 18 = 45.1 g, not 36. For CuSO₄·4H₂O: M = 231.6. Moles = 125/231.6 = 0.5397. Water = 4 × 0.5397 × 18 = 38.9 g. For CuSO₄·3H₂O: M = 213.6. Moles = 125/213.6 = 0.5853. Water = 3 × 0.5853 × 18 = 31.6 g. Closest to 36.0 g is CuSO₄·4H₂O (38.9 g) but the mole ratio method gives 2.00/0.5576 = 3.59 ≈ 4. Answer: CuSO₄·4H₂O.

In a double displacement reaction, 200 mL of 0.30 M Pb(NO₃)₂ is mixed with 300 mL of 0.20 M KI. The reaction is: Pb(NO₃)₂ + 2KI → PbI₂↓ + 2KNO₃. How many grams of PbI₂ precipitate form? (Molar mass PbI₂ = 461 g/mol)

Answer: 13.83 g

Moles of Pb(NO₃)₂ = 0.200 L × 0.30 mol/L = 0.060 mol. Moles of KI = 0.300 L × 0.20 mol/L = 0.060 mol. Stoichiometry requires 2 mol KI per 1 mol Pb(NO₃)₂; 0.060 mol Pb(NO₃)₂ needs 0.120 mol KI, but only 0.060 mol KI is available — KI is limiting. Moles of PbI₂ = 0.060 mol KI × (1 mol PbI₂ / 2 mol KI) = 0.030 mol. Mass = 0.030 × 461 = 13.83 g.

For the combustion of ethanol (C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O), a student burns 9.20 g of ethanol and collects 2.64 g of CO₂. What is the percent yield of CO₂? (Molar masses: C₂H₅OH = 46.0, CO₂ = 44.0)

Answer: 15.0%

Moles of C₂H₅OH = 9.20/46.0 = 0.200 mol. Theoretical CO₂ = 0.200 mol × 2 = 0.400 mol × 44.0 g/mol = 17.6 g. Percent yield = (2.64/17.6) × 100 = 15.0%.