Ta có: \(n_{C_2H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PT: \(2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\)
Theo PT: \(\left\{{}\begin{matrix}n_{CO_2}=2n_{C_2H_2}=0,3\left(mol\right)\\n_{H_2O}=n_{C_2H_2}=0,15\left(mol\right)\end{matrix}\right.\)
⇒ VCO2 = 0,3.22,4 = 6,72 (l)
VH2O = 0,15.22,4 = 3,36 (l)