Gas Laws using Stoichometry

Question

Hydrazine, N₂H₄, reacts with oxygen to form nitrogen gas and water.

N2H4(aq)+O2(g)⟶N2(g)+2H2O(l)

If 3.85 g of N₂H₄ reacts with excess oxygen and produces 0.650 L of N₂, at 295 K and 1.00 atm what is the percent yield of the reaction?

Answer 1

N₂H₄(aq) + O₂(g) ==> N₂(g) + 2H₂O(l) ... balanced equation

To find % yield, we must first find the theoretical yield of N₂ gas:

3.85 g N₂H₄ x 1 mol N₂H₄ / 32.0 g = 0.120 moles N₂H₄

0.120 moles N₂H₄ x 1 mol N₂ / mol N₂H₄ = 0.120 moles N₂ = theoretical yield

Next, we find the actual yield of N₂ in moles:

PV = nRT

n = PV/RT = (1.00 atm)(0.650 L) / (0.0821 Latm/Kmol)(295K)

n = 0.0268 moles N₂

% yield:

Percent yield = actual yield / theoretical yield (x100%)

Percent yield = 0.0268 mols N₂ / 0.120 mols N₂ (x100%) = 22.4% yield