Hỗn hợp khí X gồm N2 và H2 có tỉ khối so với He bằng 1,8

Ta có: $\frac{n_X}{n_Y}=\frac{{\overline{M}}_Y}{{\overline{M}}_X}=\frac{4.2}{4.1,8}=\frac{10}{9}$

Giả sử n_X = 10 mol, n_Y = 9 mol

$\frac{n_{N_2}}{n_{H_2}}=\frac{4.1,8-2}{28-4.1,8}=\frac{1}{4}$ → $n_{N_2}=2mol;n_{H_2}=8mol$

 → Hiệu suất tính theo N₂.$\begin{array}{l}{N}_2+3H_2\rightleftarrows 2N{H}_3\\ Ban\text{\hspace{0.17em}dau\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}80mol\\ \text{Phan\hspace{0.17em}ung\hspace{0.17em}\hspace{0.17em}x\hspace{0.17em}\hspace{0.17em}}\to 3x2xmol\\ \text{Sau\hspace{0.17em}pu\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2-x\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}8-3x\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2x\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}mol}\end{array}$

 → 2 – x + 8 – 3x + 2x = 9

 → x = 0,5

Hiệu suất của phản ứng là: $H=\frac{0,5}{2}.100\%=25\%$

Đáp án C