Question

Determine the limiting reactant and the mass (in g) of N2 that can be formed from 50.0g N2O4 and 45.0 g N2H4. N2O4 molar mass =92.02g/mol; N2H4 molar mass= 32.05 g/mol.

N2O4(I) +2 N2H4(I)= 3N2(g) +4 H2O(g)

Answer 1

Molar mass of N2O4= 92.02 g/mol

mass(N2O4)= 50.0 g

number of mol of N2O4,

n = mass of N2O4/molar mass of N2O4

=(50.0 g)/(92.02 g/mol)

= 0.5434 mol

Molar mass of N2H4 = 32.052 g/mol

mass(N2H4)= 45.0 g

number of mol of N2H4,

n = mass of N2H4/molar mass of N2H4

=(45.0 g)/(32.05 g/mol)

= 1.404 mol

Balanced chemical equation is:

N2O4 + 2 N2H4 ---> 3 N2 + 4 H2O

1 mol of N2O4 reacts with 2 mol of N2H4

for 0.54336 mol of N2O4, 1.08672 mol of N2H4 is required

But we have 1.403969 mol of N2H4

so, N2O4 is limiting reagent

we will use N2O4 in further calculation

Molar mass of N2 = 28.02 g/mol

According to balanced equation

mol of N2 formed = (3/1)* moles of N2O4

= (3/1)*0.54336

= 1.63008 mol

mass of N2 = number of mol * molar mass

= 1.63*28.02

= 45.7 g

Answer:

N2O4 is limiting reagent

mass of N2 formed = 45.7 g