Question

15.0 g copper (II) oxide (MM = 79.55 g/mol) reacts with an excess of ammonia (MM = 17.03 g/mol) to produce nitrogen gas, copper, and water according to the UNBALANCED reaction below. If this reaction is known to have a 72.0 % yield, what mass of nitrogen (MM = 28.01 g/mol) will be actually produced?
NH3(g) +      CuO(s) →      N2(g) +      Cu(s) +      H2O(g)

Answer 1

2NH3(g) +      3CuO(s) →      N2(g) +      3Cu(s) +      3H2O(g)

3 moles of CuO react with excess of NH3 to gives 1 moles of N2

3*79.55g of CuO react with excess of NH3 to gives 28.01g of N2

15g of CuO react with excess of NH3 to gives = 28.01*15/3*79.55    = 1.76g of N2

The theoretical yield of N2 = 1.76g

percent yield   = actual yield *100/theoretical yield

72                  = actual yield *100/1.76

actual yield = 72*1.76/100   = 1.2672g