Answer to Question #307529 in Chemistry for ave

Question

when iron is exposed to oxygen (air), it rusts. how many grams of iron are needed in order to prepare 12.25 grams of Fe2O3?

Answer 1

4Fe + 3O₂ = 2Fe₂O₃

According to the equation, n (Fe) = 4/2 x n (Fe₂O₃)

n = m / M

M (Fe₂O₃) = 159.7 g/mol

M (Fe) = 55.8 g/mol

n (Fe₂O₃) = m / M = 12.25 / 159.7 = 0.08 mol

n (Fe) = 4/2 x n (Fe₂O₃) = 4/2 x 0.08 = 0.16 mol

m (Fe) = n x M = 0.16 x 55.8 = 8.9 g