N2 (g) + 3H2 (g) -------------------------> 2NH3 (g)
(3.4 mol)
Considering the stoichiometry,
1.
nN2nH2=133.4nH2=13nH2=10.2 mol\qquad\qquad \begin {aligned} \frac{n_{N_2}}{n_{H_2}} &= \frac{1}{3}\\ \\ \frac{3.4}{n_{H_2}} &= \frac{1}{3}\\ \\ n_{H_2} &= \bold{10.2\, mol} \end {aligned}nH2nN2nH23.4nH2=31=31=10.2mol
2.
nN2nNH3=123.4nNH3=12nNH3=6.8 mol\qquad\qquad \begin {aligned} \frac{n_{N_2}}{n_{NH_3}} &= \frac{1}{2}\\ \\ \frac{3.4}{n_{NH_3}} &= \frac{1}{2}\\ \\ n_{NH_3} &= \bold{6.8\, mol} \end{aligned}nNH3nN2nNH33.4nNH3=21=21=6.8mol
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