How many grams of rust (Fe2O3) would be produced by the complete reaction of 0.15 grams of iron metal?
4 Fe + 3 O2 -> 2 Fe2O3
- Roger the MoleLv 72 months ago
(0.15 g Fe) / (55.8450 g Fe/mol) x (2 mol Fe2O3 / 4 mol Fe) x
(159.6882 g Fe2O3/mol) = 0.21 g Fe2O3
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