Một hỗn hợp X gồm một ankan và một ankin. Đốt cháy hoàn toàn hỗn...

$X\left\{ \begin{aligned}
& {{C}_{n}}{{H}_{2n+2}}\left( a \right) \\
& {{C}_{m}}{{H}_{2m-2}}\left( b \right) \\
\end{aligned} \right.\xrightarrow{{{O}_{2}}:1,15\left( mol \right)}\left\{ \begin{aligned}
& C{{O}_{2}}:{{n}_{C{{O}_{2}}}}=\dfrac{8}{3}{{n}_{X}} \\
& {{H}_{2}}O:0,7\left( mol \right) \\
\end{aligned} \right.$
$5,5\left( g \right)X\xrightarrow{AgN{{O}_{3}}/N{{H}_{3}}}14,7\left( g \right)\downarrow $
$\xrightarrow{BTNT\left( O \right)}2{{n}_{{{O}_{2}}}}=2{{n}_{C{{O}_{2}}}}+{{n}_{{{H}_{2}}O}}\to {{n}_{C{{O}_{2}}}}=0,8\left( mol \right)$
$\to {{n}_{X}}=\dfrac{3}{8}.0,8=0,3$
$\to \left\{ \begin{aligned}
& a+b=0,3 \\
& b-a=0,1 \\
\end{aligned} \right.\to \left\{ \begin{aligned}
& a=0,1 \\
& b=0,2 \\
\end{aligned} \right.\to 0,1n+0,2m=0,8\to \left[ \begin{aligned}
& \left\{ \begin{aligned}
& n=2 \\
& m=3 \\
\end{aligned} \right. \\
& \left\{ \begin{aligned}
& n=4 \\
& m=2 \\
\end{aligned} \right. \\
\end{aligned} \right.$
TH1: $\left\{ \begin{aligned}
& n=2 \\
& m=3 \\
\end{aligned} \right.\to \left\{ \begin{aligned}
& {{C}_{2}}{{H}_{6}}:0,1 \\
& {{C}_{3}}{{H}_{4}}:0,2 \\
\end{aligned} \right.\to {{m}_{1}}=0,1.30+0,2.40=11\left( gam \right)$
$\to 5,5\left( gam \right)X\left\{ \begin{aligned}
& {{C}_{2}}{{H}_{6}}:0,05 \\
& {{C}_{3}}{{H}_{4}}:0,1 \\
\end{aligned} \right.\to {{m}_{{{C}_{3}}{{H}_{3}}Ag}}=0,1.147=14,7\left( gam \right)\left( TM \right)$
TH2: $\left\{ \begin{aligned}
& n=4 \\
& m=2 \\
\end{aligned} \right.\to \left\{ \begin{aligned}
& {{C}_{4}}{{H}_{10}}:0,1 \\
& {{C}_{2}}{{H}_{2}}:0,2 \\
\end{aligned} \right.\to 5,5\left( gam \right)X\left\{ \begin{aligned}
& {{C}_{4}}{{H}_{10}}:0,05 \\
& {{C}_{2}}{{H}_{2}}:0,1 \\
\end{aligned} \right.\to {{m}_{{{C}_{2}}A{{g}_{2}}}}=0,1.240=24\left( gam \right)\left( L \right)$