chemistry question

Question

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

N2H4(aq)+O2(g)⟶N2(g)+2H2O(l)

If 4.05 g of N2H4 reacts with excess oxygen and produces 0.350 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

Since there is excess O2, we need only concern ourselves with the moles of N₂H₄ present. This will determine the theoretical yield of N₂ gas.

molar mass N₂H₄ = 32.0 g / mole

moles N₂H₄ present = 4.05 g x 1 mol / 32.0 g = 0.1266 mols

theoretical mols of N₂ = 0.1266 mols N₂H₄ x 1 mol N₂ / mol N₂H₄ = 0.1266 mol N₂

% yield = actual yield / theoretical yield (x100%). We have the theoretical yield in moles and the actual yield is given in liters. We need to convert one of the other so the units are the same. We will do it both ways using the fact that at STP (standard temperature and pressure) 1 mole of gas = 22.4 L

0.350 L x 1 mol / 22.4 L = 0.0156 moles

% yield = 0.0156 mols / 0.1266 mols (x100) = 12.3%

0.1266 mols N₂ x 22.4 L / mol = 2.84 L

0.350 L / 2.84 L (x100%) = 12.3%