a.

Xét $1(mol)$ hỗn hợp khí

Ta có hệ pt:

$\begin{cases} n_{hh}=n_{N_2}+n_{H_2}=1(mol)\\m_{hh}=28.n_{N_2}+2.n_{H_2}=2.12,4.1=24,8(g) \end{cases}\\\begin{cases} n_{N_2}=0,877(mol)\\n_{H_2}=0,123(mol) \end{cases}$

$\%V_{N_2}=\%n_{N_2}=\dfrac{0,877}{1}.100\%=87,7\%$

$\%V_{H_2}=100\%-87,7\%=12,3\%$

b.

Ta có hệ pt:

$\begin{cases} m_{hh}=28.n_{N_2}+2.n_{H_2}=6,2(g)\\n_{hh}=n_{N_2}+n_{H_2}=\dfrac{6,2}{12,4.2}=0,25(mol)\end{cases}\\\begin{cases} n_{N_2}=0,22(mol)\\n_{H_2}=0,03(mol) \end{cases}$

$V_{N_2}(đkc)=0,219.24,79=5,42901(l)$

$V_{H_2}(đkc)=0,031.24,79=0,76849(l)$

$m_{N_2}=0,219.28=6,132(g)$

$m_{H_2}=6,2-6,132=0,068(g)$

c.

Phản ứng xảy ra:

$N_2+3H_2\xrightarrow[]{xt,t^o}2NH_3$

Xét thấy $:\dfrac{0,219}{1} > \dfrac{0,03}{3}$ nên $N_2$ dư

$n_{NH_3}=\dfrac{2}{3}.n_{H_2}=0,02(mol)$

$H=85\% \to V_{NH_3}(đkc)=0,02.24,79.85\%=0,42143(l)$