Question

Nitric acid reacts violently with copper metal to produce copper (II) nitrate and a noxious brown gas, NO2. If 8.421 g of copper react with excess nitric acid and the percent yield of NO2 is 71.55% how many grams of NO2 were formed?

4 HNO3+Cu --> Cu(NO3)2+ 2 NO2+ 2 H2O

Answer 1

4 HNO3+Cu --> Cu(NO3)2+ 2 NO2+ 2 H2O

1 mole of Cu react with excess nitric acid to gives 2 moles of NO2

63.5g of Cu react with excess nitric acid to gives 2*46g of NO2

8.421g of Cu react with excess nitric acid = 2*46*8.421/63.5   = 12.2g of NO2

Theoritical yield of No2 = 12.2g

percentage yield   = actual yield*100/theoritical yield

71.55                     = x*100/12.2

x                             = 71.55*12.2/100   = 8.73g of NO2

actual yield of NO2 = 8.73g >>>>>answer