$n_P=\dfrac{3,1}{31}=0,1(mol)$
$n_{O_2}=\dfrac{1,12}{22,4}=0,05(mol)$
$4P+5O_2\xrightarrow{{t^o}} 2P_2O_5$
$\dfrac{0,1}{4}>\dfrac{0,05}{5}\to P$ dư, $O_2$ hết
$n_{P_2O_5}=\dfrac{2}{5}n_{O_2}=0,02(mol)$
$\to m_{P_2O_5}=0,02.142=2,84g$
$n_{P\rm dư}=0,1-0,02.2=0,06(mol)$
$\to m_{P\rm dư}=0,06.31=1,86g$