\(n_{H_2}=\dfrac{2,688}{22,4}=0,12mol\)
\(2CH_3COOH+Zn\rightarrow\left(CH_3COO\right)_2Zn+H_2\)
0,24 0,12 0,12 ( mol )
\(m_{CH_3COOH}=0,3.60=14,4g\)
\(\rightarrow\left\{{}\begin{matrix}\%m_{CH_3COOH}=\dfrac{14,4}{30}.100=48\%\\\%m_{C_2H_5OH}=100\%-48\%=52\%\end{matrix}\right.\)
\(m_{\left(CH_3COO\right)_2Zn}=0,12.183=21,96g\)
$ a) PTHH:\\ 2CH_3COOH + Zn \to (CH_3COO)_2Zn + H_2 \\ b) n_{H_2} = \dfrac{2,688}{22,4} = 0,12(mol) \\$
$ \to n_{CH_3COOH} = 2n_{H_2} = 0,24(mol) $
`%m_{CH_3COOH} = {0,24.60}/{30}.100% = 48%`
` %m_{C_2H_5OH} = 100 - 48 = 52%`
` n_{(CH_3COO)_2Zn} = n_{H_2} = 0,12(mol) `
` \to m_{(CH_3COO)_2Zn} = 0,12.183 = 21,96(gam)`
