How to find the percent yield and what is it?

Question

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

N2H4(aq)+O2(g)⟶N2(g)+2H2O(l)

If 2.35 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?



percent yield:

Answer 1

Percent Yield is the quantity of N2 in this case produced from the reaction divided by the quantity of N2 that should be produced (theoretical) using N2H4. The yield is typically calculated based on the mass ratio. In this case you use stoichiometric calculations to caculate the theoretical N2 that should be produced and use the ideal gas law to calculate the actual N2 that was produced.

Answer 2

N₂H₄ + O₂ ==> N₂ + 2H₂O ... balanced equation

The % yield is the actual yield divided by the theoretical yield (x100%).

So, first we need to find the theoretical yield.

To find the theoretical yield we use the mols of N₂H₄ and the balanced equation to find the theoretical yield of N₂, the product.

mols N₂H₄ used = 2.35 g x 1 mol / 32 g = 0.0734 mols

mols N₂ that can be produced = 0.0734 mol N₂H₄ x 1 mol N₂ / mol N₂H₄ = 0.0734 mols N₂ (theoretical)

mols N₂ actually formed = ?

Use PV = nRT and solve for n (moles)

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

n = 0.0268 mols N₂ produced

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

% yield = 0.0268 mol / 0.0734 mol (x100%) = 36.6% yield