Consider the balanced reaction between Cu and nitric acid shown below. If 1.55 g of Cu (NO3)2
(MW = 187.6 g/mol) was obtained from the reaction of 0.95 g of Cu (MW = 63.55 g/mol) with
excess HNO3, what is the percent yield of the reaction?
Cu(s) + 4HNO3(aq) Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l)