$a,PTPƯ:4P+5O_2\xrightarrow{t^o} 2P_2O_5$
$n_{P}=\dfrac{6,2}{31}=0,2mol.$
$n_{O_2}=\dfrac{3,36}{22,4}=0,15mol.$
$\text{Lập tỉ lệ:}$ $\dfrac{0,2}{4}>\dfrac{0,15}{5}$
$⇒P$ $dư.$
$⇒n_{P}(dư)=0,2-\dfrac{0,15.4}{5}=0,08mol.$
$⇒m_{P}(dư)=0,08.31=2,48g.$
$b,V_{O_2}=20\%V_{kk}$
$⇒V_{kk}=\dfrac{V_{O_2}}{20\%}=\dfrac{3,36}{20\%}=16,8l.$
chúc bạn học tốt!