Step 1
The reaction involved:
2 Al (s) + 3 CuSO4 (aq) => Al2(SO4)3 (aq) + 3 Cu (s) (completed and balanced9
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Step 2
Information provided:
1.25 g Al
3.28 g CuSO4
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Information needed:
The molar masses of:
Al) 27.0 g/mol
CuSO4) 160 g/mol
Al2(SO4)3) 342 g/mol
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Step 3
The limiting reactant:
By stoichiometry:
2 Al (s) + 3 CuSO4 (aq) => Al2(SO4)3 (aq) + 3 Cu (s)
2 x 27.0 g Al ------ 3 x 160 g CuSO4
1.25 g Al ------ X
X = 1.25 g Al x 3 x 160 g CuSO4/2 x 27.0 g Al
X = 11.1 g CuSO4
For 1.25 g of Al, 11.1 g of CuSO4 is needed but there is 3.28 g, so the limiting reactant is CuSO4.
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Step 4
The amount of Al2(SO4)3:
2 Al (s) + 3 CuSO4 (aq) => Al2(SO4)3 (aq) + 3 Cu (s)
3 x 160 g CuSO4 ------------ 342 g Al2(SO4)3
3.28 g CuSO4 ------------ X
X = 2.34 g
Answer: 2.34 g Al2(SO4)3