CTTQ của amin: \(C_xH_yN_t\)
PTHH: \(C_xH_yN_t+\left(x+\dfrac{y}{4}\right)O_2\xrightarrow[]{t^o}xCO_2+\dfrac{y}{2}H_2O+\dfrac{t}{2}N_2\)
\(n_{H_2O}=\dfrac{12,6}{18}=0,7\left(mol\right);n_{CO_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
Theo BTNT O: \(2n_{O_2}=2n_{CO_2}+n_{H_2O}\)
=> \(n_{O_2}=\dfrac{0,4.2+0,7}{2}=0,75\left(mol\right)\)
=> VO2 = 0,75.22,4 = 16,8 (l)
