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 vapor. If a sample containing 18.1 g of NH3 is reacted with 90.4 g of CuO, which is the limiting reactant? How many grams of N2 will be formed?

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