Câu 2:
$n_{CO_2}=\dfrac{49,5}{44}=1,125(mol)$
$n_{H_2O}=\dfrac{14,85}{18}=0,825(mol)$
$\to n_{\text{ankin}}=n_{CO_2}-n_{H_2O}=0,3(mol)$
Đặt CTTQ hai ankin là $C_nH_{2n-2}$
$\to n=\dfrac{n_{CO_2}}{n_{\text{ankin}}}=3,75$
$3<3,75<4$. Vậy CTPT 2 ankin là $C_3H_4, C_4H_6$
Câu 3:
$n_{CO_2}=\dfrac{4,62}{44}=0,105(mol)$
$n_{H_2O}=\dfrac{2,7}{18}=0,15(mol)$
$\to n_{\text{ankan}}=n_{H_2O}-n_{CO_2}=0,045(mol)$
Đặt CTTQ hai ankan là $C_nH_{2n+2}$
$\to n=\dfrac{n_{CO_2}}{n_{\text{ankan}}}=2,33$
$2<2,33<3$. Vậy CTPT 2 ankan là $C_2H_6, C_3H_8$