Đáp án:

\(\begin{array}{l}
\rm{a)\quad CH_3COOH + C_2H_5OH\quad \mathop{\overset{t^\circ}\leftrightharpoons}\limits_{H_2SO_4}\quad CH_3COOC_2H_5 + H_2O\\
b)\quad m_{CH_3COOH} = 12\, gam\\
c)\quad m_{CH_3COOC_2H_5}  = 17,6\, gam}
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
\text{a) Phương trình hóa học:}\\
\rm{CH_3COOH \quad + \quad C_2H_5OH\quad \mathop{\overset{t^\circ}\leftrightharpoons}\limits_{H_2SO_4}\quad CH_3COOC_2H_5\quad + \quad H_2O\\
b)\quad n_{CH_3COOH} = n_{C_2H_5OH} = \dfrac{m}{M} = \dfrac{9,2}{46} = 0,2\, mol\\
\Rightarrow m_{CH_3COOH} = n \times M = 0,2 \times 60 = 12\, gam\\
c)\quad n_{CH_3COOC_2H_5} = n_{C_2H_5OH} = 0,2\,mol\\
\Rightarrow m_{CH_3COOC_2H_5} = n \times M = 0,2 \times 88 = 17,6\, gam
}
\end{array}\) 

$m_{C_{2}H_{5}OH} = 9,2 (g) \\ n_{C_{2}H_{5}OH} = \dfrac{m}{M} = \dfrac{9,2}{46} = 0,2 (mol)\\ a)C_{2}H_{5}OH + CH_{3}COOH \xrightarrow{t^o, H_{2}SO_{4}} CH_{3}COOC_{2}H_{5} + H_{2}O \\ \text{Tỷ lệ mol các chất : 0,2 : 0,2 : 0,2 : 0,2 } \\ b) m_{CH_{3}COOH} = n . M = 0,2 . 60 = 12(g) \\ c) m_{CH_{3}COOC_{2}H_{5}} = n . M = 0,2 . 88 = 17,6 (g) $