Giải
nC2H4=\(\frac{V}{22,4}=\frac{3,36}{22,4}=0,15\left(mol\right)\)
PTHH: 3O2 + C2H4 -> 2CO2 + 2 H2O
P/ư : 0,45 <- 0,15 -> 0,3 (mol)
Vậy a) \(V_{O2\left(\text{đ}ktc\right)}=n.22,4=0,45.22,4=10,08\left(l\right)\)
b) \(V_{CO2\left(\text{đ}ktc\right)}=n.22,4=0,3.22,4=6,72\left(l\right)\)