Question

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

N2H4(aq) + O2(g) -> N2(g) + 2H2O(l)

If 3.45 g of N2H4 reacts and produces 0.350 L of N2, at 295 K and 1.00 atm, what is the percent yield of the reaction?

Answer 1

moles of N2H4 mass / molar mass = 3.45 g / 32.045 g/mol = 0.107 mol

from the balanced equation

1 mole of N2H4 produces 1 mol of N2 accordingly

0.107 mol of N2H4 produces 0.107 mol of N2

mass of N2 = mol of N2 x molar mass of N2 = 0.107 mol x 28.0134 g = 3 g

find the moles of N2 using PV = nRT formuls

1.0 atm x 0.35 L = n x 0.0821 L atm mol-1 K-1 x 295K

n = 0.35 / 24.2195mol-1 = 0.01445 mol

mass = 0.01445 mol x 28.0134 g/mol = 0.4048 g

% of yield = (actual mass / theritical mass] x 100

= [ 0.4048 g / 3 g] x 100

= 13.5 %