Question

When 4.50 L of hydrogen gas react with an excess of nitrogen gas at standard temperature and pressure, how many liters of ammonia gas will be produced?

Answer 1

3H₂ + N₂ = 2NH₃

V(H₂)/3 = V(NH₃)/2

V(NH₃)=2V(H₂)/3

V(NH₃)=2*4.50/3=3.00 L

3.00 L

Answer 2

Answer: The volume of ammonia gas produced will be 3 Liters.

Explanation:

At STP:

1 mole of a gas occupies 22.4 L of volume.

The chemical reaction for the formation of ammonia follows the equation:

$N_2+3H_2\rightarrow 2NH_3$

It is given that nitrogen gas is present in excess. Thus, hydrogen gas is considered as limiting reagent.

By Stoichiometry of the reaction:

$3\times 22.4L$ of hydrogen produces $2\times 22.4L$ of ammonia gas.

So, 4.5 L of hydrogen gas will produce = $\frac{2\times 22.4L}{3\times 22.4L}\times 4.5L=3L$ of ammonia gas.

Hence, the volume of ammonia gas produced will be 3 Liters.