Practice question
Question
How many grams of MnO₂ are required to produce 1.42 g of Cl₂ with excess HCl? (Molar masses: MnO₂ = 87 g/mol, Cl₂ = 71 g/mol)
Explanation
Reaction: MnO₂ + 4HCl → MnCl₂ + 2H₂O + Cl₂. Moles of Cl₂ = 1.42/71 = 0.02 mol. 1 mol Cl₂ needs 1 mol MnO₂; 0.02 mol needs 0.02 × 87 = 1.74 g.
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