Chemical Reactions and Stoichiometry Flashcards

In the reaction 2KMnO₄ + 16HCl → 2KCl + 2MnCl₂ + 5Cl₂ + 8H₂O, if 31.6 g of KMnO₄ (MW = 158 g/mol) reacts with excess HCl, how many grams of Cl₂ (MW = 71 g/mol) are produced?

Answer: 17.75 g

Moles of KMnO₄ = 31.6 ÷ 158 = 0.200 mol. The balanced equation shows 2 mol KMnO₄ produces 5 mol Cl₂ (ratio 5:2). Moles of Cl₂ = 0.200 × (5/2) = 0.250 mol. Mass = 0.250 × 71 = 17.75 g.

A student performs a synthesis and obtains 38.7 g of product. After reviewing the procedure, they discover a math error that caused them to underestimate the theoretical yield — the correct theoretical yield is 50.0 g, not the 45.0 g originally calculated. What is the corrected percent yield?

Answer: 77.4%

Percent yield = (actual yield ÷ theoretical yield) × 100. The actual yield (38.7 g, measured in the lab) is unchanged by the calculation correction. Using the corrected theoretical yield: (38.7 ÷ 50.0) × 100 = 77.4%. The originally calculated value of 45.0 g is simply wrong and discarded.

For the reaction N₂ + 3H₂ → 2NH₃, if 28 g of N₂ and 9 g of H₂ are mixed, which statement correctly identifies the limiting reagent and the mass of excess reagent remaining?

Answer: N₂ is limiting; 3.0 g of H₂ remains unused

Moles of N₂ = 28 ÷ 28 = 1.00 mol; moles of H₂ = 9 ÷ 2 = 4.50 mol. The stoichiometric ratio requires 3 mol H₂ per mol N₂. To consume 1.00 mol N₂ you need 3.00 mol H₂ — you have 4.50, so N₂ is limiting. Excess H₂ = 4.50 − 3.00 = 1.50 mol × 2 g/mol = 3.0 g.

Which of the following reactions is best classified as a disproportionation reaction?

Answer: Cl₂ + 2NaOH → NaCl + NaOCl + H₂O

A disproportionation reaction occurs when a single element is simultaneously oxidized and reduced. In Cl₂ + 2NaOH → NaCl + NaOCl + H₂O, chlorine starts at oxidation state 0 and ends at −1 in NaCl (reduced) and +1 in NaOCl (oxidized). The same element is both oxidized and reduced, which is the hallmark of disproportionation.

Combustion of an unknown hydrocarbon CₓHᵧ produces 8.8 g of CO₂ and 5.4 g of H₂O. What is the empirical formula of the hydrocarbon?

Answer: CH₃

Each mole of CO₂ contains 1 mol C: 8.8 ÷ 44 = 0.200 mol C. Each mole of H₂O contains 2 mol H: (5.4 ÷ 18) × 2 = 0.600 mol H. Simplest ratio C:H = 0.200 : 0.600 = 1 : 3. Empirical formula = CH₃.

40.0 mL of 0.50 M Pb(NO₃)₂ is mixed with 150 mL of 0.80 M KI. A precipitate of PbI₂ forms (MW = 461 g/mol). What is the maximum mass of PbI₂ that can be produced?

Answer: 9.22 g

Moles of Pb²⁺ = 0.0400 L × 0.50 mol/L = 0.020 mol. Moles of I⁻ = 0.150 L × 0.80 mol/L = 0.120 mol. The precipitation reaction is Pb²⁺ + 2I⁻ → PbI₂, requiring a 1:2 ratio. To consume 0.020 mol Pb²⁺, only 0.040 mol I⁻ is needed — far less than the 0.120 mol available, so Pb²⁺ is the limiting reagent. Moles PbI₂ = 0.020 mol; mass = 0.020 × 461 = 9.22 g.