$a,PTPƯ:4P+5O_2\xrightarrow{t^o} 2P_2O_5$
$n_{P}=\dfrac{7,75}{31}=0,25mol.$
$n_{O_2}=\dfrac{8,96}{22,4}=0,4mol.$
$\text{Lập tỉ lệ:}$ $\dfrac{0,25}{4}<\dfrac{0,4}{5}$
$⇒O_2$ 4dư.$
$⇒n_{O_2}(dư)=0,4-\dfrac{0,25.5}{4}=0,0875mol.$
$⇒m_{O_2}(dư)=0,0875.32=2,8g.$
$b,Theo$ $pt:$ $n_{P_2O_5}=\dfrac{1}{2}n_{P}=0,125mol.$
$⇒m_{P_2O_5}=0,125.142=17,75g.$
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