P/S: mình giải theo 6,2g P
PTHH : 4P + 5O2 $→$ 2P2O5
$n_P=$ $\dfrac{6,2}{31}=0,2(mol)$
$a)$ Theo PTHH $n_{P_2O_5}=0,5\times n_P=0,5\times 0,2=0,1(mol)$
$→m_{P_2O_5}=0,1\times 142=14,2(g)$
$b)$ Theo PTHH $n_{O_2}=1,25\times n_P= 1,25\times 0,2=0,25(mol)$
$→V_{O_2}=0,25\times 22,4=5,6(l)$
$c)$ PTHH: 2KMnO4 $→$ K2MnO4 + MnO2 + O2
Theo PTHH $→$ $n_{KMnO_4}=2\times n_{O_2}=2 \times 0,25=0,5(mol)$
$→m_{KMnO_4}=0,5\times 158 =79(g)$