Answer to Question #109814 in General Chemistry for Rashed Ahmed idris

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