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Câu hỏi: Cracking 4,48 lít butan (đktc), thu được hỗn hợp X gồm 6 chất H2, CH4, C2H6, C2H4, C3H6, C4H8. Dẫn hết hỗn hợp X vào bình dung dịch brom dư thì thấy khối lượng bình brom tăng 8,4 gam và bay ra khỏi bình brom là hỗn hợp khí Y. Thể tích oxi (đktc) cần đốt hết hỗn hợp Y là
A. 5,6 lít.
B. 8,96 lít.
C. 4,48 lít.
D. 6,72 lít.
A. 5,6 lít.
B. 8,96 lít.
C. 4,48 lít.
D. 6,72 lít.
+ Sơ đồ phản ứng: ${{C}_{4}}{{H}_{10}}\xrightarrow{cracking}\left\{ \begin{aligned}
& {{H}_{2}}+{{C}_{4}}{{H}_{8}} \\
& C{{H}_{4}}+{{C}_{3}}{{H}_{6}} \\
& {{C}_{2}}{{H}_{6}}+{{C}_{2}}{{H}_{4}} \\
\end{aligned} \right\}\xrightarrow{+B{{r}_{2}}}\left\{ \begin{aligned}
& \left\{ \begin{aligned}
& {{C}_{4}}{{H}_{8}}B{{r}_{2}} \\
& {{C}_{3}}{{H}_{6}}B{{r}_{2}} \\
& {{C}_{2}}{{H}_{4}}B{{r}_{2}} \\
\end{aligned} \right\} \\
& \left\{ \begin{aligned}
& {{H}_{2}},C{{H}_{4}} \\
& {{C}_{2}}{{H}_{6}} \\
\end{aligned} \right\}\xrightarrow{{{O}_{2}},t{}^\circ }\left\{ \begin{aligned}
& C{{O}_{2}} \\
& {{H}_{2}}O \\
\end{aligned} \right\} \\
\end{aligned} \right.$
+ $\left\{ \begin{aligned}
& {{n}_{\left( {{H}_{2}},C{{H}_{4}},{{C}_{2}}{{H}_{6}} \right)}}={{n}_{\left( {{C}_{4}}{{H}_{8}},{{C}_{3}}{{H}_{6}},{{C}_{2}}{{H}_{4}} \right)}}={{n}_{{{C}_{4}}{{H}_{10}}\ bd}}=0,2 \\
& {{m}_{\left( {{C}_{4}}{{H}_{8}},{{C}_{3}}{{H}_{6}},{{C}_{2}}{{H}_{4}} \right)}}={{m}_{\text{bình B}{{\text{r}}_{2}}\text{ tăng}}}=8,4 \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& {{\overline{M}}_{\left( {{C}_{4}}{{H}_{8}},{{C}_{3}}{{H}_{6}},{{C}_{2}}{{H}_{4}} \right)}}=42 \\
& {{\overline{M}}_{\left( {{H}_{2}},C{{H}_{4}},{{C}_{2}}{{H}_{6}} \right)}}=16={{M}_{C{{H}_{4}}}} \\
\end{aligned} \right.$
+ $\left\{ \begin{aligned}
& {{\overline{M}}_{Y}}=16\Leftrightarrow Y\text{ la C}{{\text{H}}_{4}} \\
& 8\underbrace{{{n}_{C{{H}_{4}}}}}_{0,2}=4\underbrace{{{n}_{{{O}_{2}}}}}_{?} \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& {{n}_{{{O}_{2}}}}=0,4 mol \\
& {{V}_{{{O}_{2}}}}=8,96 lit \\
\end{aligned} \right.$
& {{H}_{2}}+{{C}_{4}}{{H}_{8}} \\
& C{{H}_{4}}+{{C}_{3}}{{H}_{6}} \\
& {{C}_{2}}{{H}_{6}}+{{C}_{2}}{{H}_{4}} \\
\end{aligned} \right\}\xrightarrow{+B{{r}_{2}}}\left\{ \begin{aligned}
& \left\{ \begin{aligned}
& {{C}_{4}}{{H}_{8}}B{{r}_{2}} \\
& {{C}_{3}}{{H}_{6}}B{{r}_{2}} \\
& {{C}_{2}}{{H}_{4}}B{{r}_{2}} \\
\end{aligned} \right\} \\
& \left\{ \begin{aligned}
& {{H}_{2}},C{{H}_{4}} \\
& {{C}_{2}}{{H}_{6}} \\
\end{aligned} \right\}\xrightarrow{{{O}_{2}},t{}^\circ }\left\{ \begin{aligned}
& C{{O}_{2}} \\
& {{H}_{2}}O \\
\end{aligned} \right\} \\
\end{aligned} \right.$
+ $\left\{ \begin{aligned}
& {{n}_{\left( {{H}_{2}},C{{H}_{4}},{{C}_{2}}{{H}_{6}} \right)}}={{n}_{\left( {{C}_{4}}{{H}_{8}},{{C}_{3}}{{H}_{6}},{{C}_{2}}{{H}_{4}} \right)}}={{n}_{{{C}_{4}}{{H}_{10}}\ bd}}=0,2 \\
& {{m}_{\left( {{C}_{4}}{{H}_{8}},{{C}_{3}}{{H}_{6}},{{C}_{2}}{{H}_{4}} \right)}}={{m}_{\text{bình B}{{\text{r}}_{2}}\text{ tăng}}}=8,4 \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& {{\overline{M}}_{\left( {{C}_{4}}{{H}_{8}},{{C}_{3}}{{H}_{6}},{{C}_{2}}{{H}_{4}} \right)}}=42 \\
& {{\overline{M}}_{\left( {{H}_{2}},C{{H}_{4}},{{C}_{2}}{{H}_{6}} \right)}}=16={{M}_{C{{H}_{4}}}} \\
\end{aligned} \right.$
+ $\left\{ \begin{aligned}
& {{\overline{M}}_{Y}}=16\Leftrightarrow Y\text{ la C}{{\text{H}}_{4}} \\
& 8\underbrace{{{n}_{C{{H}_{4}}}}}_{0,2}=4\underbrace{{{n}_{{{O}_{2}}}}}_{?} \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& {{n}_{{{O}_{2}}}}=0,4 mol \\
& {{V}_{{{O}_{2}}}}=8,96 lit \\
\end{aligned} \right.$
Đáp án B.
