Gọi $n_{O_{2}}$ = x (mol)
$n_{N_{2}}$ = y (mol)
$m_{X}$ = $m_{O_{2}}$ + $m_{N_{2}}$ = 32x + 28y = 2.32 (g) (1)
Do $d_{X/H_{2}}$ = 14.5 ⇒ $M_{X}$ = 14.5 × 2 = 29 (g)
⇒ $n_{X}$ = 2.32 ÷ 29 = 0.08 (mol) ⇒ x + y = 0.08 (2)
Từ (1) và (2) ⇒ $\left \{ {{x=0.02} \atop {y=0.06}} \right.$
$n_{S}$ = $\frac{0.32}{32}$ = 0.01 (mol)
Có Pư:
$S_{}$ + $O_{2}$ → $SO_{2}$
Bđ: 0.01 0.02 0 (mol)
Pư: 0.01->0.01->0.01 (mol)
Sau: 0 0.01 0.01 (mol)
⇒$M_{Y}$ = $\frac{0.01× 32 + 0.01 × 64 + 0.06 × 28}{0.01 + 0.01 + 0.06}$ = 33 (g)
⇒ $d_{Y/H_{2}}$ = 33 ÷ 2 = 16.5
