$n_P=\dfrac{1,55}{31}=0,05(mol)$
PTHH (1): $4P+5O_2\xrightarrow{t^o}2P_2O_5$
PTHH: (2) $2KMnO_4\xrightarrow{t^o}MnO_2+O_2+K_2MnO_4$
Theo PTHH (1):
$n_{O_2}=\dfrac{5}{4}n_P=\dfrac{5}{4}\cdot0,05=0,0625(mol)$
$V_{O_2}=0,0625.22,4=1,4(l)$
Theo PTHH (2):
$n_{KMnO_4}=2n_{O_2}=2.0,0625=0,125(mol)$
Khối lượng $KMnO_4$ cần dùng là:
$m_{KMnO_4}=0,125.158=19,75(g)$