a) PTHH: \(Mg+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Mg+H_2\)
b) Ta có: \(n_{Mg}=\frac{2,4}{24}=0,1\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}n_{\left(CH_3COO\right)_2Mg}=0,1mol\\n_{H_2}=0,1mol\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{\left(CH_3COO\right)_2Mg}=0,1\cdot142=14,2\left(g\right)\\V_{H_2}=0,1\cdot22,4=2,24\left(l\right)\end{matrix}\right.\)
c) PTHH: \(CH_3COOH+C_2H_5OH\rightarrow CH_3COOC_2H_5+H_2O\)
Ta có: \(n_{CH_3COOH}=0,2mol\) \(\Rightarrow n_{C_2H_5OH}=0,2mol\)
\(\Rightarrow m_{C_2H_5OH}=0,2\cdot46=9,2\left(g\right)\) (Lý thuyết)
Mặt khác: \(V_{C_2H_5OH}=23\cdot40\%=9,2\left(ml\right)\) \(\Rightarrow m_{C_2H_5OH}=9,2\cdot0,8=7,36\left(g\right)\) (Thực tế)
\(\Rightarrow H\%=\frac{7,36}{9,2}\cdot100=80\%\)
nMg=\(\frac{m}{M}\)=\(\frac{2.4}{12}=0.2\)
PTHH: 2CH3COOH+Mg --->(CH3COO)2Mg +H2
mCH3COO)2Mg=n.M=0.2.130=26(g)
VH2=0.2.22.4=4.48(l)
c)Vrượu=\(\frac{40.23}{100}=9.2\)(ml)
mC2H5OH=D.V=0,8.9,6=7.68(g)
PTHH C2H5OH+CH3COOH--> CH3COOC2H5+H2O
mC2H5OH=0,2.46=9,2(g)
Hiệu suất=\(\frac{7.68}{9.2}.100\%\)=83,5%
Mình ko dám chắc mik lm đúng nhưng chúc bạn học tốt