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.

Step-by-step explanation:

Given values:

Number of molecules of ammonia gas =
`7.5* 10^(26)`

According to the mole concept:

`6.022* 10^(23)` number of molecules are present in 1 mole of a compound

So,
`7.5* 10^(26)` number of molecules will be present in
`(1)/(6.022* 10^(23))* 7.5* 10^(26)=1.245* 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* 10^3` moles of ammonia gas will be produced from
`(1)/(2)* 1.245 * 10^3=6.225* 10^2` moles of nitrogen gas

At STP conditions:

1 mole of a gas occupies 22.4 L of volume

So,
`6.225 * 10^2` moles of nitrogen gas will occupy
`(22.4 L)/(1 mol)* 6.225* 10^2 moles=13944 L` of volume

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