Đáp án $+$ Giải thích các bước giải:

$a,$

$M_X=0,775×16=12,4(g/mol)$

$⇒\dfrac{n_{N_2}}{n_{H_2}}=\dfrac{12,4-2}{28-12,4}=\dfrac{2}{3}$

$→\%V_{N_2\ (X)}=\dfrac{2}{2+3}×100\%=40\%$

$→\%V_{H_2\ (X)}=100-40=60\%$

$M_Y=1,25×16=20(g/mol)$

Giả sử có $2\ mol\ N_2$ và $3\ mol\ H_2$

$N_2+3H_2→2NH_3$

Xét $\dfrac{2}{1}>\dfrac{3}{3}⇒$ Tính theo $H_2$

Đặt $n_{H_2\ (pư)}=x(mol)$

$⇒n_{H_2\ (dư)}=3-x(mol)$

Theo PT:

$n_{N_2\ (pư)}=\dfrac{x}{3}(mol)$

$⇒n_{N_2\ (dư)}=2-\dfrac{x}{3}$

$n_{NH_3}=\dfrac{2}{3}x(mol)$

$M_Y=\dfrac{28.(2-\dfrac{x}{3})+2.(3-x)+17×\dfrac{2}{3}x}{(2-\dfrac{x}{3}+(3-x)+\dfrac{2}{3}x}=20(g/mol)$

$⇒x=2,85$

$n_Y=(2-\dfrac{2,85}{3}+(3-2,85)+\dfrac{2}{3}×2,85=3,1(mol)$

$→\%V_{NH_3\ (Y)}=\dfrac{\dfrac{2}{3}×2,85}{3,1}×100\%≈61,29\%$

$→\%V_{H_2\ (Y)}=\dfrac{3-2,85}{3,1}×100\%≈4,83\%$

$→\%V_{N_2\ (Y)}=100-61,29-4,83=33,88\%$

$b,$

$→H\%=\dfrac{2,85}{3}×100\%=95\%$