Answer to Question #114467 in General Chemistry for mike

Question

Nitrogen gas can be prepared by passing gaseous ammonia over solid copper (II) oxide at high temperatures. The other products of the reaction are solid copper and water vapour. If 18.10g of NH3 is reacted with 90.40g of CuO,
(i). Explain, with justification, which is the limiting reactant?
(ii).How many grams of N2 will be formed?

Answer 1

3 CuO + 2 NH3 → 3 Cu + N2 + 3 H2O;

(i). n(CuO)=m(CuO)/(3*Mr(CuO))=90.40/(3*80)=0.376 mol

n(NH3)=m(NH3)/(2*Mr(NH3))=18.10/(2*17)=0.532 mol

n(CuO)<n(NH3), so CuO is the limiting reactant;

(ii). m(N2)=m(CuO)*Mr(N2)/(3*Mr(CuO))=90.40*28/(3*80)=10.546 g

Answer: (i). CuO is the limiting reactant;

(ii). 10.546 grams of N2 will be formed