Answer to Question #170198 in General Chemistry for josh

Question

Nitrogen and hydrogen combine at a high temperature, in the presence of a catalyst, to produce ammonia.

N

2

(g)+3H

2

(g)⟶2NH

3

(g)

N2(g)+3H2(g)⟶2NH3(g)

Assume 0.280 mol N

2

0.280 mol N2 and 0.894 mol H

2

0.894 mol H2 are present initially.

After complete reaction, how many moles of ammonia are produced?

Answer 1

Solution:

As is clear from the chemical equation, the molar ratio $\frac{H_2} {N_2} =\frac{3 }{1}$.

$\frac{n(H_2) } {n(N_2) } =\frac{0.46 }{0.14}=3.286$ i.e. H₂ was taken in excess and some part of it remains unreacted. N₂ s limiting reactant, it reacts completely and 0 moles of N₂ remains unreacted.

Number of moles of NH₂ produced may be calculated from the proportion:

 1 mol (N₂) – 2 mol (NH₃) (according to the chemical equation)

0.140 mol (N₂) – X mol (NH₃)

X = 0.140 · 2 / 1 = 0.280 mol of ammonia are produced.