Ta có:VO2=$\frac{1}{5}$ Vkk
⇒V kk=5.V O2
a.
n C=$\frac{24}{12}$ =2 (mol)
C+O2$\xrightarrow{t^o} $CO2↑
2→ 2 (mol)
V O2(đktc)=2.22,4=44,8 (l)
Vkk=5.44,8=224 (l)
b.
n P=$\frac{6,2}{31}$=0,2 (mol)
4P+5O2→2P2O5
0,2→0,25 (mol)
V O2(đktc)=0,25.22,4=5,6 (l)
Vkk=5.5,6=28 (l)
-----------------Nguyễn Hoạt-----------------