Cho 8,8 gam C2H5COOCH3...

A

- Hướng dẫn giải

Phương pháp giải:

Giải chi tiết:

Đáp án A

Ta có : \(n({C_2}{H_5}{\text{COOC}}{{\text{H}}_3}) = 0,1;\,n(KOH) = 0,12.\)

\(\begin{gathered}
{C_2}{H_5}{\text{COOC}}{{\text{H}}_3} + KOH\xrightarrow{{{t^0}}}{C_2}{H_5}{\text{COOK + C}}{{\text{H}}_3}OH \hfill \\
\,\,\,0,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,1 \hfill \\
\end{gathered} \)

BTLK ta có 8,8  + 56.0,12= m + 32.0,1; suy ra m= 12,32 gam.