$PTHH: 4P + 5O_{2}\quad \underrightarrow{t^{o}}\quad 2P_{2}O_{5}$
$a)n_{P}=\frac{6,2}{31}=0,2 (mol)$
$n_{O_{2}}=\frac{11,2}{22,4}=0,5 (mol)$
$\text{Vì }\frac{n_{P}}{4}<\frac{n_{O_{2}}}{5}$
$\Rightarrow O_{2}\text{ còn dư}$
$\text{Theo ptpư: }n_{P_{2}O_{5}}=\frac{1}{2}n_{P}=0,1 (mol)$
$\Rightarrow m_{P_{2}O_{5}}=0,1.142=14,2 (g)$
$b) \text{Theo ptpư: }n_{O_{2}}=\frac{5}{4}n_{P}=0,25 (mol)$
$\Rightarrow V_{O_{2}}=0,25.22,4=5,6 (l)$
$c) 2KMnO_{4}\quad\underrightarrow{t^{o}}\quad{K_{2}MnO_{4}}+MnO_{2}+O_{2}\uparrow$
$\text{Theo ptpư: }n_{KMnO_{4}}=2n_{O_{2}}=1 (mol)$
$\Rightarrow m_{KMnO_{4}}=1.158=158 (g)$