Đáp án đúng: A
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.