Question

4. When 1.00 g of N2 is mixed with 2.25 g of H2 according to the following balanced equation, how many grams of N2 are left over? N2(g) + 3 H2(g) + 2NH3(g) Molar Masses: 28.02 g/mol 2.016 g/mol 342.150 g/mol

Answer 1

Molar mass of N2 = 28.02 g/mol

mass(N2)= 1.0 g

use:

number of mol of N2,

n = mass of N2/molar mass of N2

=(1 g)/(28.02 g/mol)

= 3.569*10^-2 mol

Molar mass of H2 = 2.016 g/mol

mass(H2)= 2.25 g

use:

number of mol of H2,

n = mass of H2/molar mass of H2

=(2.25 g)/(2.016 g/mol)

= 1.116 mol

Balanced chemical equation is:

N2 + 3 H2 ---> 2 NH3

1 mol of N2 reacts with 3 mol of H2

for 3.569*10^-2 mol of N2, 0.1071 mol of H2 is required

But we have 1.116 mol of H2

so, N2 is limiting reagent

Since N2 is limiting reagent, there will not be any N2 left over

Answer: 0 g