\(\)\(X\left\{\begin{matrix} CH_3COOC_2H_5\\ C_2H_3COOH\\ CH_3CHO & \end{matrix}\right. + \overset{O_2( t^{o})}{\rightarrow} \left\{\begin{matrix} CO_2 \\ H_2O \end{matrix}\right. + \overset{Ca(OH)_2}{\rightarrow} CaCO_3\)Theo đề bài:
\(n_{CaCO_3}\)= 0,45 (mol)
BTC---> \(n_{CO_2} = n_{CaCO_3}\)= 0,45 (mol)
Khối lượng bình nước vôi tăng 27g
→\(m_{CO_2}+ m_{H_2O}=27 \)
→\(0,45.44+ 18.n_{H_2O} =27\)→ \(\)\(n_{H_2O}=0,4 (mol)
\)\(\)
\(n_{CO_2}-n_{H_2O} = n_{C_2H_3COOH}\\
\rightarrow n_{C_2H_3COOH} = 0,45-0,4= 0,05 (mol)\)