Gọi số mol $O_{2}:x$
$⇒nO_{\text{trong oxit}}=2x$
$KL+Cl_{2} \to \text{Muối clorua}(1)$
$KL+O_{2} \to Oxit $
$Oxit+HCl \to \text{Muối clorua}(2)+H_{2}O$
2x 4x
$2H^{+}+O^{2-} \to H_{2}O$
4x 2x
Ta có:
$mMuối=mKL+mCl_{2}+mCl_\text{trong HCl} \\ ⇔49,95=18+71nCl_{2}+4x.35,5 \\ ⇒71nCl_{2}+142x=31,95\\71nCl_2+142nO_2=31,95(1)$
Ta có :
$\frac{nCl_{2}}{nO_{2}}=\frac{1}{4} \\ ⇒4nCl_{2}-nO_{2}=0(2) \\ (1)(2)⇒\left \{ {{nCl_{2}=0,05} \atop {nO_{2}=0,2}} \right. \\ V=VCl_{2}+VO_{2}=0,05.22,4+0,2.22,4=5,6lit$