In a reaction, 41.8 g of chromium(III) oxide reacts with 14.8 g of aluminum to produce chromium and aluminum oxide. If 28.6 g of chromium is produced, what mass of aluminum oxide is produced? How do I go about solving this problem?

Answer 1

It's actually easier than it seems. Just requires simple stoichiometry.

Balanced Equation:
Cr₂O₃ + 2 Al -> Al₂O₃ + 2 Cr

28.6 g Cr * (1 mol Cr / 52.00 g Cr) * (1 mol Al₂O₃ / 2 mol Cr) * (101.96 g Al₂O₃/ 1 mol Al₂O₃) = 28.0 g Al₂O₃ (with sig figs)