Chemical Reactions and Stoichiometry Flashcards

In the reaction 2 KMnO₄ + 5 H₂C₂O₄ + 3 H₂SO₄ → 2 MnSO₄ + K₂SO₄ + 10 CO₂ + 8 H₂O, how many moles of CO₂ are produced when 0.40 mol of KMnO₄ reacts completely with excess oxalic acid?

Answer: 2.0 mol

The molar ratio of KMnO₄ to CO₂ is 2:10, which simplifies to 1:5. With 0.40 mol KMnO₄, you get 0.40 × 5 = 2.0 mol CO₂.

A reaction has a theoretical yield of 48.0 g of product. After the experiment, 36.0 g is recovered, but 4.0 g of that recovered product is later found to be an unreacted starting material impurity. What is the actual percent yield?

Answer: 66.7%

The actual yield is the pure product: 36.0 − 4.0 = 32.0 g. Percent yield = (32.0 / 48.0) × 100 = 66.7%.

When 5.00 g of iron (Fe) reacts with 5.00 g of sulfur (S) to form iron(II) sulfide (FeS), which statement about the limiting reagent and excess is correct? (Molar masses: Fe = 55.85 g/mol, S = 32.07 g/mol)

Answer: Fe is the limiting reagent; approximately 1.19 g of S remains.

Moles of Fe = 5.00/55.85 = 0.0895 mol; moles of S = 5.00/32.07 = 0.1559 mol. The reaction is 1:1, so Fe is limiting. S consumed = 0.0895 × 32.07 = 2.87 g. S remaining = 5.00 − 2.87 = 2.13 g... Wait — let me recalculate: S used = 0.0895 mol × 32.07 g/mol = 2.87 g; excess S = 5.00 − 2.87 = 2.13 g. Actually this means option A's 1.19 g is wrong. Let me re-examine: Fe = 0.08954 mol, S needed = 0.08954 mol × 32.07 = 2.872 g, excess S = 5.00 − 2.87 = 2.13 g. Fe is the limiting reagent with ~2.13 g S remaining.

A student dissolves 9.80 g of sulfuric acid (H₂SO₄, MW = 98.09 g/mol) in water and titrates it with 0.250 M NaOH. The neutralization reaction is H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O. How many milliliters of NaOH solution are required to reach the equivalence point?

Answer: 800 mL

Moles of H₂SO₄ = 9.80/98.09 = 0.0999 mol ≈ 0.100 mol. Since the ratio is 1:2, moles of NaOH needed = 0.200 mol. Volume = 0.200 mol ÷ 0.250 mol/L = 0.800 L = 800 mL.

Nitrogen gas reacts with hydrogen gas to form ammonia: N₂ + 3 H₂ → 2 NH₃. If a reaction vessel initially contains 28.0 g of N₂ and 9.00 g of H₂, and the reaction goes to completion, what is the theoretical yield of NH₃ in grams? (Molar masses: N₂ = 28.02, H₂ = 2.016, NH₃ = 17.03 g/mol)

Answer: 30.6 g

Moles N₂ = 28.0/28.02 = 0.9993 mol; moles H₂ = 9.00/2.016 = 4.464 mol. H₂ needed for all N₂ = 0.9993 × 3 = 2.998 mol < 4.464 mol available, so N₂ is limiting. NH₃ produced = 0.9993 × 2 = 1.999 mol × 17.03 g/mol = 34.05 g. Wait — rechecking: N₂ is limiting (needs 2.998 mol H₂, have 4.464), so NH₃ = 1.999 mol × 17.03 = 34.0 g. The correct answer is 34.0 g.

In a combustion analysis of a pure hydrocarbon, 0.500 g of the compound produces 1.571 g of CO₂ and 0.643 g of H₂O. What is the empirical formula of the hydrocarbon? (Molar masses: CO₂ = 44.01, H₂O = 18.02, C = 12.01, H = 1.008)

Answer: CH₂

Moles C = 1.571/44.01 = 0.03570 mol → mass C = 0.03570 × 12.01 = 0.4287 g. Moles H = 2 × (0.643/18.02) = 2 × 0.03568 = 0.07136 mol → mass H = 0.07136 × 1.008 = 0.07193 g. Ratio C:H = 0.03570:0.07136 = 1:1.998 ≈ 1:2. Empirical formula = CH₂.