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What volume of O2 gas (in L), measured at 762mmHg and 16∘C, is required to completely react with 50.1g of Al?
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- Roger the MoleLv 77 years agoFavorite Answer
4 Al + 3 O2 → 2 Al2O3
(50.1 g Al) / (26.98154 g Al/mol) x (3 mol O2 / 4 mol Al) = 1.3926 mol O2
V = nRT / P = (1.3926 mol) x (62.36367 L mmHg/K mol) x (16 + 273 K) / (762 mmHg) = 32.9 L O2
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