Đáp án + Giải thích các bước giải:

$1)$

$n_{CO_2}=0,3(mol);n_{H_2O}=0,3(mol)$

Bảo toàn C,H: $n_C=n_{CO_2}=0,3(mol);n_H=2n_{H_2O}=0,6(mol)$

$\Rightarrow n_O=\dfrac{7,4-0,3.12-0,6}{16}=0,2(mol)$

$\Rightarrow X$ gồm $C,H,O$

$b)M_X=2,3125.32=74(g/mol)$

Đặt CTPT là $C_xH_yO_z$

$\Rightarrow x:y:z=0,3:0,6:0,2=3:6:2$

$\Rightarrow CTPT$ có dạng $(C_3H_6O_2)_n$

$\Rightarrow (12.3+6+16.2).n=74$

$\Rightarrow n=1$

$\Rightarrow CTPT:C_3H_6O_2$

$2)$

$a)n_{N_2}+n_{O_2}=n_X=\dfrac{0,672}{22,4}=0,03(1)$

Mà $m_{N_2}+m_{O_2}=0,88$

$\Rightarrow 28n_{N_2}+32n_{O_2}=0,88(2)$

Từ $(1)(2)\Rightarrow n_{N_2}=0,02(mol);n_{O_2}=0,01(mol)$

$\Rightarrow \begin{cases} \%V_{N_2}=\dfrac{0,02}{0,02+0,01}.100\%=66,67\%\\ \%V_{O_2}=100-66,67=33,33\% \end{cases}$

$b)\overline{M}_X=\dfrac{0,88}{0,03}=\dfrac{88}{3}(g/mol)$

$\Rightarrow n_X=2,2:\dfrac{88}{3}=0,075(mol)$

$\Rightarrow V_{H_2}=V_X=0,075.22,4=1,68(l)$