a) nKMnO4= 94,8/158=0,6(mol)
PTHH: 2 KMnO4 -to-> K2MnO4 + MnO2 + O2
0,6_________________0,3______0,3___0,3(mol)
V(O2,đktc)= 0,3.22,4=6,72(l)
b) PTHH: 4 P + 5 O2 -to-> 2 P2O5
nP= 6,4/31= 32/155(mol)
nO2= 0,3(mol)
Ta có : 32/155 :4 < 0,3/5
-> P hết, O2 dư, tính theo nP
mP(p.ứ)= 6,4(g)
nP2O5= 2/4 . nP= 2/4 . 32/155= 16/155(mol)
=> mP2O5= 16/155.142\(\approx14,658\left(g\right)\)