# Chemistry

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What is the percent yield if 4.65g of copper is produced when 1.87g of aluminum reacts with an excess of copper(II) sulfate?

2Al(s)+3CuSO4(aq)==>Al2(SO4)3(aq)+3Cu(s)

• Chemistry -

1.87 g Al = ?? moles.
moles = 1.87/molar mass Al.

Using the coefficients in the balanced equation, convert moles Al to moles Cu.

Now convert moles Cu to grams. g = moles x molar mass Cu. This is the theoretical yield of the reaction.

%yield = (actual yield/theoretical yield)*100 = ??
The actual yield is given in the problem as 4.65 grams.

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