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Fe2O3 + 2Al --> 2Fe + Al2O3

Before:

50.0g + 50.0g > 0.0g + (not important)

159.69(g/mol) + 26.98(g/mol)

In mol (before reaction):

+0.3131 mol + 1.853 mol (excess)

Reaction (used reactant > formed Fe):

-0.3131 mol - 0.6262 mol > + 0.6262 mol Fe

Remaining (= before - used):

0.0 mol Fe2O3 + 1.227 mol Al > 0.6262 mol Fe, this should be multiplied by the molar mass of Fe to get mass in grams: 0.6262 (molFe) * 55.85 (g/molFe) = 34.97 = 35.0 g Fe

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13y ago

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