Question

determine how many grams of N2 are produced from the reaction of 9.70 g of H2O2 and 5.08 g of N2H4.

Answer 1

Molar mass of N2H4,
MM = 2*MM(N) + 4*MM(H)
= 2*14.01 + 4*1.008
= 32.052 g/mol


mass(N2H4)= 5.08 g

number of mol of N2H4,
n = mass of N2H4/molar mass of N2H4
=(5.08 g)/(32.052 g/mol)
= 0.1585 mol

Molar mass of H2O2,
MM = 2*MM(H) + 2*MM(O)
= 2*1.008 + 2*16.0
= 34.016 g/mol


mass(H2O2)= 9.7 g

number of mol of H2O2,
n = mass of H2O2/molar mass of H2O2
=(9.7 g)/(34.016 g/mol)
= 0.2852 mol
Balanced chemical equation is:
N2H4 + 2 H2O2 ---> N2 + 4 H2O


1 mol of N2H4 reacts with 2 mol of H2O2
for 0.1585 mol of N2H4, 0.317 mol of H2O2 is required
But we have 0.2852 mol of H2O2

so, H2O2 is limiting reagent
we will use H2O2 in further calculation


Molar mass of N2 = 28.02 g/mol

According to balanced equation
mol of N2 formed = (1/2)* moles of H2O2
= (1/2)*0.2852
= 0.1426 mol


mass of N2 = number of mol * molar mass
= 0.1426*28.02
= 4.00 g

Answer: 4.00 g