Assuming that you can only make 863 g Ag, what is the percent yield of Ag if you actually recover 1018 g Ag?

1 Answer
May 15, 2015

#"percent-yield"=("actual yield")/("theoretical yield")xx100#

If I read your question correctly, the theoretical yield is #"863 g"# and the actual yield is #"1018 g"#.

#"percent yield"=("1018 g")/("863 g")xx100=118%#

This means that a greater mass of reactants was used than was required from the calculations, or it could mean that a calculation error occurred when determining the theoretical yield.

It is not uncommon for students to use a greater mass of the reactants than the reaction requires from their calculations, because they are afraid they won't get enough product, but it's best not to do this.