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Sagot :
Answer: The volume of nitrogen gas that is required is 13944 L.
Explanation:
Given values:
Number of molecules of ammonia gas = [tex]7.5\times 10^{26}[/tex]
According to the mole concept:
[tex]6.022\times 10^{23}[/tex] number of molecules are present in 1 mole of a compound
So, [tex]7.5\times 10^{26}[/tex] number of molecules will be present in [tex]\frac{1}{6.022\times 10^{23}}\times 7.5\times 10^{26}=1.245\times 10^3 moles[/tex] of ammonia gas
The chemical equation for the formation of ammonia gas follows:
[tex]N_2 (g) + 3H_2(g) \rightarrow 2NH_3(g)[/tex]
According to the stoichiometry of the reaction:
2 moles of ammonia gas are produced from 1 mole of nitrogen gas
So, [tex]1.245\times 10^3[/tex] moles of ammonia gas will be produced from [tex]\frac{1}{2}\times 1.245 \times 10^3=6.225\times 10^2[/tex] moles of nitrogen gas
At STP conditions:
1 mole of a gas occupies 22.4 L of volume
So, [tex]6.225 \times 10^2[/tex] moles of nitrogen gas will occupy [tex]\frac{22.4 L}{1 mol}\times 6.225\times 10^2 moles=13944 L[/tex] of volume
Hence, the volume of nitrogen gas that is required is 13944 L.
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