Question

Nitrogen gas, N_{2} can be prepared by passing gaseous ammonia over solid copper(II) oxide, CuO, at high temperature. The other products of the reaction are solid copper, Cu, and water vapor.

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

In a certain experiment, a reaction mixture containing 100.0 g CuO and excess amount of ammonia, NH 3 , yields 7.46 g N_{2} What is the percent yield for this experiment?

Answer 1

Reaction:

3CuO + 2NH₃ = N₂ + 3Cu + 3H₂O

n(CuO) = ${\frac {m(CuO)} {64+16}}={\frac {90.4} {80}}=1.13$ mol

n(NH₃) = ${\frac {m(NH_3)} {14+3}}={\frac {18.1} {17}}=1.065$ mol

For 1.065 mol of NH₃ undergoing the reaction, ${\frac {3} {2}}$ times larger amount of CuO is needed.

Needed amount of CuO is 1.6 mol but it's only 1.13 of it so CuO is a limiting reactant, we must count stoichiometry by amount of this reactant.

n(CuO) = 1.13 mol, n(N₂ formed) = ${\frac {n(CuO)} {3}}={\frac {1.13} {3}}=0.3767$ mol

m(N₂) = 0.3767*28 = 10.55 grams