What is the percent yield if 248.7 g of Cu is produced when 87 g of Al reacts with an excess of CuSO4 in a single replacement reaction?

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:

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