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Câu hỏi: 10 gam hỗn hợp X gồm metan, propen và axetilen làm mất màu 48 gam $\mathrm{Br}_{2}$ trong dung dịch. Mặt khác, 13,44 lít khí X (đktc) tác dụng vừa đủ với $\text{AgN}{{\text{O}}_{3}}\text{/N}{{\text{H}}_{\text{3}}}$ được 36 gam kết tủa. Thành phần phần trăm về khối lượng của metan có trong $X$ là
A. 42,3.
B. 54,4.
C. 37,8.
D. 44,8.
A. 42,3.
B. 54,4.
C. 37,8.
D. 44,8.
$10\left\{ \begin{array}{*{35}{l}}
\begin{aligned}
& \text{C}{{\text{H}}_{4}}:x \\
& {{\text{C}}_{2}}{{\text{H}}_{2}}:y\xrightarrow{\text{B}{{\text{r}}_{\text{2}}}} \\
\end{aligned} \\
{{\text{C}}_{2}}{{\text{H}}_{4}}:z \\
\end{array}\left\{ \begin{array}{*{35}{l}}
16x+26y+28z=10 (1) \\
2y+z=0,3\left( {{n}_{{{\text{B}}_{2}}}} \right) (2) \\
\end{array} \right. \right.$
$0,6\left\{ \begin{array}{*{35}{l}}
\begin{aligned}
& \text{C}{{\text{H}}_{4}}:x.k \\
& {{\text{C}}_{2}}{{\text{H}}_{2}}:y.k \\
\end{aligned} \\
{{\text{C}}_{2}}{{\text{H}}_{4}}:z.k \\
\end{array}\xrightarrow{\text{AgN}{{\text{O}}_{3}}/\text{N}{{\text{H}}_{3}}}{{\text{C}}_{2}}\text{A}{{\text{g}}_{2}}:36\operatorname{gam}\left\{ \begin{array}{*{35}{l}}
y.k=0,15\left( {{n}_{{{\text{C}}_{2}}\text{A}{{\text{g}}_{2}}}}={{n}_{{{\text{C}}_{2}}{{\text{H}}_{2}}}} \right) \\
xk+yk+zk=0,6 \\
\end{array}\to \dfrac{y}{x+y+z}=\dfrac{0,15}{0,6} \right. \right. \left( 3 \right)$
$\rightarrow\left\{\begin{array}{l}x=0,34 \\ y=0,128 \\ z=0,044\end{array}\right.$
$\%_{\mathrm{CH}_{4}}=\dfrac{0,34.16}{10} .100=54,4 \%$
\begin{aligned}
& \text{C}{{\text{H}}_{4}}:x \\
& {{\text{C}}_{2}}{{\text{H}}_{2}}:y\xrightarrow{\text{B}{{\text{r}}_{\text{2}}}} \\
\end{aligned} \\
{{\text{C}}_{2}}{{\text{H}}_{4}}:z \\
\end{array}\left\{ \begin{array}{*{35}{l}}
16x+26y+28z=10 (1) \\
2y+z=0,3\left( {{n}_{{{\text{B}}_{2}}}} \right) (2) \\
\end{array} \right. \right.$
$0,6\left\{ \begin{array}{*{35}{l}}
\begin{aligned}
& \text{C}{{\text{H}}_{4}}:x.k \\
& {{\text{C}}_{2}}{{\text{H}}_{2}}:y.k \\
\end{aligned} \\
{{\text{C}}_{2}}{{\text{H}}_{4}}:z.k \\
\end{array}\xrightarrow{\text{AgN}{{\text{O}}_{3}}/\text{N}{{\text{H}}_{3}}}{{\text{C}}_{2}}\text{A}{{\text{g}}_{2}}:36\operatorname{gam}\left\{ \begin{array}{*{35}{l}}
y.k=0,15\left( {{n}_{{{\text{C}}_{2}}\text{A}{{\text{g}}_{2}}}}={{n}_{{{\text{C}}_{2}}{{\text{H}}_{2}}}} \right) \\
xk+yk+zk=0,6 \\
\end{array}\to \dfrac{y}{x+y+z}=\dfrac{0,15}{0,6} \right. \right. \left( 3 \right)$
$\rightarrow\left\{\begin{array}{l}x=0,34 \\ y=0,128 \\ z=0,044\end{array}\right.$
$\%_{\mathrm{CH}_{4}}=\dfrac{0,34.16}{10} .100=54,4 \%$
Đáp án B.