N2H4(aq) + O2(g) = N2(g) + 2H2O (I)

Question

If 3.05 g of N2H4 reacts with excess oxygen and produces .750 L of N2 at 295 K and 1 atm, what is the percent yield

Answer 1

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

To find the % yield, we will need to find both the theoretical yield and the actual yield.

Theoretical yield:

3.05 g N₂H₄ x 1 mol N₂H₄/32 g x 1 mol N₂/1 mol N₂H₄ = 0.0953 moles N₂ theoretically formed

Actual yield:

Using PV = nRT and solving or n (moles of N₂), we have....

n = PV/RT = (1 atm)(0.750 L) / (0.0821 Latm/Kmol)(295K)

n = 0.031 moles N₂

Percent yield:

% = actual yield / theoretical yield (x100%) = 0.031 mol / 0.095 mol (x100%) = 32.6% yield