1. Given: 2Al(s) + Fe2O3(s) à Al2O3(s) + 2Fe(l) Step 1 is done! How many grams of Fe2O3 are needed to produce 6.7 g of Fe? Step 2: Mass to moles Step 3: Mole to Mole (Ratio) Step 4: Mole to Mass

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2Al(s) + Fe2O3(s) = Al2O3(s) + 2Fe(l)

n=m/M

M (Fe) = 55.9 g/mol

M (Fe2O3) = 159.7 g/mol

n (Fe)=6.7/55.9 = 0.12 mol

n (Fe2O3) = 1/2 x n (Fe) = 1/2 x 0.12 = 0.06 mol

m (Fe2O3) = n x M = 0.06 x 159.7 = 9.6 g

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