nKCLO3 = 24.5/122.5 = 0.2 (mol)
2KClO3 -> (to) 2KCl + 3O2
0.2 -----------------------> 0.3
VO2 = 0.3 × 22.4 = 6.72 l
nP = 3.1/31= 0.1 (mol)
4P+ 5O2 -> 2P2O5(to)
Ban đầu :0.1 : 0.3
P/ứ : 0.1 -> 0.125
Sau p/ứ : 0 : 0.175
=> mO2 dư = 0.175 × 32= 5.6 g