If 3.45 g of N2 H4 reacts with excess oxygen and produces 0.950 L of N2 at 295K and 1.00 ATM what is the percent yield of the reaction

Answer 1

N2H4 + O2 ===> N2 + 2H2O

Molar mass of N2H4 = 32.0 g/mol

mole of N2H4 = 3.45 g / 32.0 g/mol = 0.107 mol

stochiometric ration between N2H2 and N2 is 1:1

Hence during the reaction 0.107 N2 was released (theoricaly)

Now lets find the practically evolved N2 mol using PV=nRT

n=PV/RT

n= (1 atm * 0.95 L)/ (295 K* 0.082057)

N= 0.0392 mol

Yield= (0.0392/0.107)*100 = 36.68%