Answer:
7.82 g of Cu
Explanation:
2 moles of Al react to 3 moles of copper sulfate in order to produce 3 moles of copper and 1 mol of aluminum sulfate.
Firstly we determine the moles of reactant.
As copper sulfate is in excess, Al is the limiting.
2.75 g . 1mol /26.98g = 0.102 moles
Ratio is 2:3. 2 moles of Al, can produce 3 moles of Cu
So the 0.102 moles of Al will produce(0.102 . 3) /2 = 0.153 moles.
We convert moles to mass: 0.153 mol . 63.5g /mol = 9.71 g
That's the theoretical yield (100 % yield reaction)
We know that: (yield produced / theoretical yield) . 100 = percent yield
We replace:
(Yield produced / 9.71g) . 100 = 80.5 %
(Yield produced / 9.71g) = 0.805
Yield produced = 0.805 . 9.71g = 7.82 g
