$n_{C_3H_6Br_2}=\dfrac{20,4}{42+160}=0,1(mol)$

Bảo toàn $C$:

$n_{C_3H_6}=n_{C_3H_6Br_2}=0,1(mol)$

$\to \%m_{C_3H_6}=\dfrac{0,1.42.100}{7,6}=55,26\%$

$\to \%m_{CH_4}=100-55,26=44,74\%$

Đáp án:

\(\begin{array}{l}
\%m_{C_3H_6}= 55,26\%\\
\%m_{CH_4}= 44,74\%
\end{array}\) 

Giải thích các bước giải:

Sửa đề: $20,2\ g$ đibrom propan

\(\begin{array}{l}
n_{C_3H_6Br_2} = \dfrac{20,2}{202} = 0,1\ \rm mol\\
PTHH:\\
\rm C_3H_6\quad +\quad Br_2\quad \longrightarrow\quad C_3H_6Br_2\\
0,1\ \text{mol} \qquad \qquad \qquad \qquad 0,1 \ \text{mol}\\
n_{C_3H_6} = n_{C_3H_6Br_2} = 0,1\ \rm mol\\
m_{C_3H_6} = 0,1 \times 42 = 4,2\ g\\
\%m_{C_3H_6} = \dfrac{4,2\times 100\%}{7,6} = 55,26\%\\
\%m_{CH_4} = 100\% - 55,26\% = 44,74\%
\end{array}\)