Answer: The mass of [tex]N_2[/tex] required are, 198 grams.
Explanation : Given,
Mass of [tex]NH_3[/tex] = 240.0 g
Molar mass of [tex]NH_3[/tex] = 17 g/mol
First we have to calculate the moles of [tex]NH_3[/tex].
[tex]\text{Moles of }NH_3=\frac{\text{Given mass }NH_3}{\text{Molar mass }NH_3}[/tex]
[tex]\text{Moles of }NH_3=\frac{240.0g}{17g/mol}=14.12mol[/tex]
Now we have to calculate the moles of [tex]N_2[/tex]
The balanced chemical equation is:
[tex]N_2+3H_2\rightarrow 2NH_3[/tex]
From the reaction, we conclude that
As, 2 mole of [tex]NH_3[/tex] produces from 1 mole of [tex]N_2[/tex]
So, 14.12 mole of [tex]NH_3[/tex] produces form [tex]\frac{14.12}{2}=7.06[/tex] mole of [tex]N_2[/tex]
Now we have to calculate the mass of [tex]N_2[/tex]
[tex]\text{ Mass of }N_2=\text{ Moles of }N_2\times \text{ Molar mass of }N_2[/tex]
Molar mass of [tex]N_2[/tex] = 28 g/mole
[tex]\text{ Mass of }N_2=(7.06moles)\times (28g/mole)=197.68g\approx 198g[/tex]
Therefore, the mass of [tex]N_2[/tex] required are, 198 grams.
