Practice question
Question
How many grams of Al are required to produce 5.6 L of H₂ at STP with excess HCl? (Atomic mass: Al = 27)
Explanation
Reaction: 2Al + 6HCl → 2AlCl₃ + 3H₂. Moles of H₂ = 5.6/22.4 = 0.25 mol. 3 mol H₂ from 2 mol Al; 0.25 mol from (2/3)×0.25 ≈ 0.1667 mol Al. Mass = 0.1667 × 27 ≈ 4.5 g.
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