Đáp án:
Phản xảy ứng xảy ra:
\[S+O_2\to SO_2\qquad(1)\]
\[4P+5O_2\to 2P_2O_5\qquad(2)\]
$n_{SO_2}=\dfrac{4,48}{22,4}=0,2(mol)$
Theo PTHH $(1)\to n_S=n_{SO_2}=0,2(mol)$
$\to m_S=0,2.32=6,4(g)$
$\to m_P=18,8-6,4=12,4(g)$
$\to n_P=\dfrac{12,4}{31}=0,4(mol)$
Theo PTHH: $\sum n_{O_2}=n_S+\dfrac{5}{4}.n_P=0,2+\dfrac{5}{4}.0,4=0,7(mol)$
$\to m_{O_2}=0,7.32=22,4(g)$