What is the maximum amount of al2 so4 3 which could be formed from the reaction?
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2Al(s) + 3CuSO4(aq) --> Al2(SO4)3 + 3Cu(s)
12.71g ...10.09g ................? g
12.71g Al x (1 mol Al / 27.0g Al) x (1 mol Al2(SO4)3 / 2 mol Al) x (342.3g Al2(SO4)3 / 1 mol Al2(SO4)3) = 80.57 g Al2(SO4)3
10.09g CuSO4 x (1 mol CuSO4 / 159.6 g CuSO4) x (1 mol Al2(SO4)3 / 3 mol CuSO4) x (342.3g Al2(SO4)3 / 1 mol Al2(SO4)3) = 7.213g Al2(SO4)3
The actual yield for the reaction is 7.213g of Al2(SO4)3. It will be expressed to four significant figures. It also means that CuSO4 is the limiting reactant.
12.71g ...10.09g ................? g
12.71g Al x (1 mol Al / 27.0g Al) x (1 mol Al2(SO4)3 / 2 mol Al) x (342.3g Al2(SO4)3 / 1 mol Al2(SO4)3) = 80.57 g Al2(SO4)3
10.09g CuSO4 x (1 mol CuSO4 / 159.6 g CuSO4) x (1 mol Al2(SO4)3 / 3 mol CuSO4) x (342.3g Al2(SO4)3 / 1 mol Al2(SO4)3) = 7.213g Al2(SO4)3
The actual yield for the reaction is 7.213g of Al2(SO4)3. It will be expressed to four significant figures. It also means that CuSO4 is the limiting reactant.
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