Đáp án:

$m_1 = 3,6(g)$

$m_2 = 8,8(g)$

Giải thích các bước giải:

$n_X = \dfrac{2,24}{22,4} = 0,1 (mol)$

$M_X = 14.2 = 28 (g/mol)$

$\rightarrow m_X = 0,1.28 = 2,8 (g)$

Dựa theo CTHH $C_2H_2, C_2H_4, C_2H_6: n_C = 2n_X = 2.0,1 = 0,2 (mol)$

$\rightarrow m_H = 2,8-0,2.12 = 0,4 (g)$

$\rightarrow n_H = \dfrac{0,4}{1} = 0,4 (mol)$

BTNT:

$n_{H_2O} = \dfrac{1}{2} n_H = \dfrac{1}{2} . 0,4 = 0,2 (mol)$

$n_{CO_2} = n_C = 0,2 (mol)$

$\rightarrow m_1 = m_{H_2O} =0,2.18 = 3,6 (g)$

$\rightarrow m_2 = m_{CO_2} = 0,2.44 = 8,8 (g)$

$\text{#Bbi F ><}$

$n_X=\dfrac{2,24}{22,4}=0,1\ (mol)$

$\overline{M}_X=2.14=28\ (g/mol)$

$\to m_X=0,1.28=2,8\ (g)$

Quy hỗn hợp $X$ gồm $C_2H_2$, $C_2H_4$, $C_2H_6$ về $C_2H_n$

$\xrightarrow{\ \big[C\big]\ } n_{CO_2}=n_C=2n_X=0,2\ (mol)$

$\xrightarrow{\ \big[H\big]\ } n_{H_2O}=\dfrac12n_H=\dfrac12.(2,8-0,2.12)=0,2\ (mol)$

$\to \begin{cases}m_1=m_{H_2O}=0,2.18=3,6\ (g)\\ \quad \\m_2=m_{CO_2}=0,2.44=8,8\ (g)\end{cases}$