\(\left(1\right)C_2H_4+H_2O\xrightarrow[rượu]{men}C_2H_5OH\\ \left(2\right)C_2H_5OH+O_2\xrightarrow[giấm]{men}CH_3COOH+H_2O\\ \left(3\right)\left(4\right)\left(5\right)CH_3COOH+C_2H_5OH⇌_{(H_2SO_4.đặc,t^0)}CH_3COOC_2H_5+H_2O\\ \left(5\right)2CH_3COOH+Mg\xrightarrow[]{}\left(CH_3COO\right)Mg+H_2\)

\(a.Fe_3O_4+8HCl\xrightarrow[]{}FeCl_2+2FeCl_3+4H_2O\)     

\(2NaOH+FeCl_2\xrightarrow[]{}Fe\left(OH\right)_2+2NaCl\\ 3NaOH+FeCl_3\xrightarrow[]{}Fe\left(OH\right)_3+3NaCl\\ NaOH+HCl\xrightarrow[]{}NaCl+H_2O\\ 2Fe\left(OH\right)_3\xrightarrow[]{t^0}Fe_2O_3+3H_2O\\ 4Fe\left(OH\right)_2+O_2\xrightarrow[]{t^0}2Fe_2O_3+4H_2O\)

A gồm \(FeCl_2,FeCl_3,HCl.dư\)

B gồm \(NaCl\)

C gồm \(Fe\left(OH\right)_2,Fe\left(OH\right)_3\)

\(b.BTNT\left(Fe\right):3n_{Fe_3O_4}=2n_{Fe_2O_3}\\ \Leftrightarrow3.0,1=2n_{Fe_2O_3}\\ \Rightarrow n_{Fe_2O_3}=0,15mol\\ m=m_{Fe_2O_3}=0,15.160=24g\)

\(n_{Br_2}=\dfrac{16}{160}=0,1mol\\ C_2H_4+Br_2\xrightarrow[]{}C_2H_4Br_2\\ n_{C_2H_4}=n_{Br_2}=0,1mol\\ V_{C_2H_4}=0,2.22,4=4,48l\\ V_{CH_4}=11,2-4,48=6,72l\)

\(n_{H_2}=\dfrac{1,12}{22,4}=0,05mol\\ 2CH_3COOH+Zn\xrightarrow[]{}\left(CH_3COO\right)_2Zn+H_2\\ n_{CH_3COOH}=2n_{H_2}=0,1mol\\ C_{\%CH_3COOH}=\dfrac{0,1.60}{75}\cdot100\%=8\%\)

ui mik nhìn ko ra

rượu etylic mới đúng he