Question Detail
Question
70 grams of manganese dioxide is mixed with 3.5 moles of hydrochloric acid. How many grams of Cl₂ will be produced from this reaction if the yield for the process is 42%? [Mn = 55, O = 16, Cl = 35.5, H = 1] MnO₂ + 4HCl(aq) → MnCl₂ + Cl₂(g) + 2H₂O
Correct Answer
Option D is the correct answer.
Detailed Explanation
Molar mass MnO₂ = 55 + 32 = 87 g/mol. Moles MnO₂ = 70/87 = 0.8046 mol. Moles HCl = 3.5 mol. From the equation, 1 mol MnO₂ requires 4 mol HCl. For 0.8046 mol MnO₂, required HCl = 4 × 0.8046 = 3.2184 mol. Since 3.5 mol HCl is available, MnO₂ is the limiting reagent. Theoretical moles of Cl₂ from MnO₂ = 0.8046 mol. Mass of Cl₂ (theoretical) = 0.8046 mol × 71 g/mol = 57.13 g. Actual yield at 42% = 0.42 × 57.13 g = 24.0 g.
Hint
Identify the limiting reagent to find the theoretical yield of Cl₂, then apply the percentage yield.