Đáp án:
1mol chốt pho 2 96g cacbon
$+)$ 1 mol P
$4P+5O2\to 2P_2O_5$
Theo PTHH $\to n_{O_2}=\dfrac{1.5}{4}=1,25(mol)$
$\to m_{O_2}=32.1,25=40(g)$
$+)$ 2,96g cacbon
$n_C=\dfrac{2,96}{12}=\dfrac{37}{150}$
$C+O_2\to CO_2$
$n_{O_2}=n_C=\dfrac{37}{150}$
$\to m_{O_2}=\dfrac{37}{150}\times 32=7,9(g)$