$4P+5O_2\xrightarrow{t^o}2P_2O_5$
$n_P=\dfrac{8,8}{31}=\dfrac{44}{155}(mol)$
a. Theo PTHH:
$n_{O_2}=\dfrac{5}{4}n_P=\dfrac{5}{4}.\dfrac{44}{155}=\dfrac{11}{31}(mol)$
Thể tích khí oxi tham gia phản ứng (đktc) là:
$V_{O_2}=\dfrac{11}{31}.22,4≈7,948(l)$
Theo PTHH: $n_{P_2O_5}=\dfrac{1}{2}n_{P}=\dfrac{1}{2}.\dfrac{44}{155}=\dfrac{22}{155}(mol)$
Khối lượng $P_2O_5$ tạo thành sau phản ứng là:
$m_{P_2O_5}=\dfrac{22}{155}.142≈20,154(g)$
b. PTHH: $4P+5O_2\xrightarrow{t^o}2P_2O_5$
$n_P=\dfrac{44}{155}(mol)$
$n_{O_2}=\dfrac{80}{22,4}=\dfrac{25}{7}(mol)$
So sánh tỉ lệ:
$\dfrac{\dfrac{44}{155}}{4}<\dfrac{\dfrac{25}{7}}{5}$
$\Rightarrow O_2$ dư, $P$ phản ứng hết.
$n_{O_2\,\,\,p/ư}=\dfrac{11}{31}(mol)$
$n_{O_2\,\,\,dư}=\dfrac{25}{7}-\dfrac{11}{31}=\dfrac{698}{217}(mol)$
$m_{O_2\,\,\,dư}=\dfrac{698}{217}.32≈102,930(g)$