Sửa đề cho dễ làm : dd K2CO3 13,8%
PTHH: \(2CH_3COOH+K_2CO_3\rightarrow2CH_3COOK+H_2O+CO_2\uparrow\)
a+b) Ta có: \(n_{CH_3COOH}=\dfrac{150\cdot12\%}{60}=0,3\left(mol\right)\)
\(\Rightarrow n_{K_2CO_3}=n_{CO_2}=0,15\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}m_{ddK_2CO_3}=\dfrac{0,15\cdot138}{13,8\%}=150\left(g\right)\\V_{CO_2}=0,15\cdot22,4=3,36\left(l\right)\end{matrix}\right.\)
c) Theo PTHH:: \(n_{CH_3COOK}=0,3\left(mol\right)\) \(\Rightarrow m_{CH_3COOK}=0,3\cdot98=29,4\left(g\right)\)
Mặt khác: \(m_{CO_2}=0,15\cdot44=6,6\left(g\right)\)
\(\Rightarrow m_{dd}=m_{ddCH_3COOH}+m_{ddK_2CO_3}-m_{CO_2}=293,4\left(g\right)\)
\(\Rightarrow C\%_{CH_3COOK}=\dfrac{29,4}{293,4}\cdot100\%\approx10,02\%\)