Question

Ammonia (NH3) is produced in the Haber process by passing nitrogen (N2) with hydrogen over an iron catalyst at high temperature and pressure.

N2(g) + 3 H2(g) --> 2 NH3(g)

How many grams of ammonia can be prepared by reaction of 9.405 g of nitrogen with 2.413 g of hydrogen?

Answer 1

Molar mass of N2 = 28.02 g/mol


mass(N2)= 9.405 g

use:
number of mol of N2,
n = mass of N2/molar mass of N2
=(9.405 g)/(28.02 g/mol)
= 0.3357 mol

Molar mass of H2 = 2.016 g/mol


mass(H2)= 2.413 g

use:
number of mol of H2,
n = mass of H2/molar mass of H2
=(2.413 g)/(2.016 g/mol)
= 1.197 mol
Balanced chemical equation is:
N2 + 3 H2 ---> 2 NH3 +


1 mol of N2 reacts with 3 mol of H2
for 0.3357 mol of N2, 1.007 mol of H2 is required
But we have 1.197 mol of H2

so, N2 is limiting reagent
we will use N2 in further calculation


Molar mass of NH3,
MM = 1*MM(N) + 3*MM(H)
= 1*14.01 + 3*1.008
= 17.034 g/mol

According to balanced equation
mol of NH3 formed = (2/1)* moles of N2
= (2/1)*0.3357
= 0.6713 mol


use:
mass of NH3 = number of mol * molar mass
= 0.6713*17.03
= 11.4 g

Answer: 11.4 g