Question

Hydrazine, N2H4, may react with oxygen to form nitrogen gas and water. N2H4(aq) + O2(g)→N2(g) + 2H2O(l). If 2.05 g of N2H4 reacts and produces 0.750 L of N2, at 295 K and 1.00 atm, what is the percent yield of the reaction?

Answer 1

Given,

N2H4(aq) + O2(g)→N2(g) + 2H2O(l)

Mass of N2H4 = 2.05 g

Molar Mass of N2H4 = 32.045 g / mol

=> Moles of N2H4 = 2.05 / 32.045 = 0.064 moles

Volume of N2 = 7.5 L

Temp. = 295 K

Pressure = 1 atm

Calculating number of moles of N2 produced using PV = nRT

=> 1 x 0.75 = n x 0.0821 x 295

=> n = 0.031 moles

According to the stoichiometry of the given reaction 1 mole of N2H4 reacts to prduce 1 mole of N2

Therefore ideally (theoretically) 0.064 moles of N2H4 should produce 0.064 moles of N2

=> % yield = Actual x 100 / Theoretical = 0.031 x 100 / 0.064 = 48.44 %