Question

How many liters of nitrogen gas (N₂) are required to produce 7.5 x 10²⁶ molec NH₃ gas with excess of hydrogen gas at STP?

Answer 1

Answer: The volume of nitrogen gas that is required is 13944 L.

Explanation:

Given values:

Number of molecules of ammonia gas = $7.5\times 10^{26}$

According to the mole concept:

$6.022\times 10^{23}$ number of molecules are present in 1 mole of a compound

So, $7.5\times 10^{26}$ number of molecules will be present in $\frac{1}{6.022\times 10^{23}}\times 7.5\times 10^{26}=1.245\times 10^3 moles$ of ammonia gas

The chemical equation for the formation of ammonia gas follows:

$N_2 (g) + 3H_2(g) \rightarrow 2NH_3(g)$

According to the stoichiometry of the reaction:

2 moles of ammonia gas are produced from 1 mole of nitrogen gas

So, $1.245\times 10^3$ moles of ammonia gas will be produced from $\frac{1}{2}\times 1.245 \times 10^3=6.225\times 10^2$ moles of nitrogen gas

At STP conditions:

1 mole of a gas occupies 22.4 L of volume

So, $6.225 \times 10^2$ moles of nitrogen gas will occupy $\frac{22.4 L}{1 mol}\times 6.225\times 10^2 moles=13944 L$ of volume

Hence, the volume of nitrogen gas that is required is 13944 L.