Practice question
Question
How many grams of AlCl₃ are required to produce 5.4 g of Al with excess HCl? (Molar masses: AlCl₃ = 133.5 g/mol, Al = 27 g/mol)
Explanation
Moles of Al = 5.4/27 = 0.2 mol. 2 mol Al from 2 mol AlCl₃; 0.2 mol from 0.2 mol. Mass = 0.2 × 133.5 = 26.7 g.
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