Step 1
The reaction must be written, completed, and balanced:
2Al(s) + Fe2O3(aq) => Al2O3(aq) + 2Fe(s)
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Step 2
Information provided:
The mass of Al = 129.63 g
The mass of Fe2O3 = 767.44 g
(because of this, the limiting reactant must be determined)
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Information needed:
The molar masses:
Al) 26.981 g/mol
Fe2O3) 159.69 g/mol
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Step 3
The limiting reactant:
2Al(s) + Fe2O3(aq) => Al2O3(aq) + 2Fe(s)
By stoichiometry => 1 mole Al = 26.981 g and 1 mole Fe2O3 = 159.69 g
Procedure:
2 x 26.981 g Al --------- 159.69 g Fe2O3
129.63 g Al ---------- X
X = 129.63 g Al x 159.69 g Fe2O3/2 x 26.981 g Al = 383.61 g
(For 129.63 g of Al, 383.61 g of Fe2O3 is needed, but there is 767.44 g of Fe2O3. Therefore, the limiting reactant is the Al and the excess is the Fe2O3)
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Step 4
The mass of Al2O3: (molar mass of Al2O3 = 101.96 g/mol)
2Al(s) + Fe2O3(aq) => Al2O3(aq) + 2Fe(s)
2 x 26.981 g Al ------- 101.96 g Al2O3
129.63 g Al ------- X = 244.93 g
Answer: mass of aluminum oxide = 244.93 g