Assuming an efficiency of 47.30%, calculate the actual yield of magnesium nitrate formed from 119.6 g of magnesium and excess copper(II) nitrate. Mg+Cu(NO3)2⟶Mg(NO3)2+Cu

Question

Assuming an efficiency of  calculate the actual yield of magnesium nitrate formed from  of magnesium and excess copper(II) nitrate.

Answer 1

Mg+ Cu(NO3)2 ==> Mg(NO3)2 + Cu ... balanced equation

Since Cu(NO3)2 is in excess, the Mg will determine the theoretical yield.

119.6 g Mg x 1 mol Mg/24.3 g x 1 mol Mg(NO3)2 / 1 mol Mg x 148 g Mg(NO3)2/mol = 728.4 g Mg(NO3)2

At an efficiency of 47.30%, the actual yield would be...

728.4 g x 0.4730 = 344.5 g