Hỗn hợp X gồm ancol metylic, etylen glicol. Cho m gam X phản ứng với Na dư, thu được 2,24 lít H2 (đktc).

* Hướng dẫn giải

Đáp án A

\(\begin{align} & C{{H}_{3}}-OH+Na\to C{{H}_{3}}-ONa+1/2{{H}_{2}} \\ & \,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x/2 \\ \end{align}\)

\(\begin{align} & {{C}_{2}}{{H}_{4}}{{(OH)}_{2}}+2Na\to {{C}_{2}}{{H}_{4}}{{(ONa)}_{2}}+{{H}_{2}} \\ & \,\,\,\,\,\,\,\,\,\,y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y \\ \end{align}\)

Theo phương trình phản ứng trên ta có: \(\frac{x}{2}+y={{n}_{{{H}_{2}}}}=0,1\,(mol)\)

Mặt khác \(\xrightarrow{BTC}{{n}_{C{{O}_{2}}}}=x+2y=2\underbrace{\left( \frac{x}{2}+y \right)}_{{{n}_{{{H}_{2}}}}}=2.0,1=0,2\,(mol)\to a=8,8\,(gam)\)