posted by Keith on .
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?
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.