Cho $O_{2\text{..pứ}}=x(mol)$
Do $O_2$ thừa, mà $n_{O_2\text{..pứ}}=n_{S\text{pứ}}$
$\to n_S$ dư.
$n_S=\dfrac{12}{32}=0,375(mol)$
$\to n_S, n_{O_2}$ dư $0,375-x(mol)$
Phương trình:
$S+O_2\xrightarrow{t^o}SO_2$
Ta có: $\overline{M_{\text{hh khí}}}=28.2=56g/mol$
$\overline{M_{\text{hh khí}}}=\dfrac{{m_{SO_2}}+m_{O_{\text{dư}}}}{n_{SO_{2\text{tt}}}+n_{O_2\text{dư}}}$
$\to 56=\dfrac{64x+32(0,375-x)}{0,375-x+x}$
$\to 56=\dfrac{32x+12}{0,375}$
$\to x=\dfrac{9}{32}(mol)$
$\to n_{O_2}=x=\dfrac{9}{32}(mol)$
$\to V_{O_2\text{pứ}}=\dfrac{9}{32}.22,4=6,3(l)$
