Answer:
2.09 g Cu
Explanation:
First, we have to balance the equation:
3Al(s) + 3CuSO4(aq) = Al2(SO4)3(aq) + 3Cu(s)
Then we have to change grams of Al to mol:
1 mol Al = 26.98 g
1.37 g Al x (1 mol/26.98 g) = 0.051 mol
Then, we use the balanced equation to find the mol of Cu produced:
According to the equation:
3 mol Al produces 3 mol of Cu
3 mol Al = 3 mol Cu
So:
0.051 mol Al x (3 mol Cu/3 mol Al) = 0.051 mol Cu
This mol value is the theoric yield of the reaction.
Remember the equation for yield percent:
%Y = (Y real / Y theoric) x 100
Y theoric = 0.051 mol Cu
% Y = 67.4%
Let's replace these values on the equation and then fin Y real.
67.4 % = (Y real / 0.051) x 100
Y real = 0.033 mol Cu
So the real produced mol is 0.033 mol Cu.
Let's change these moles to grams:
1 mol Cu = 63.55 g
0.033 mol Cu x ( 63.55 g Cu/ 1 mol Cu) = 2.09 g Cu
If you want, watch the image attached is more clear the
mathematical procedure
Explanation:
