1) 

a) \(2{C_2}{H_5}OH + 2K\xrightarrow{{}}2{C_2}{H_5}OK + {H_2}\)

b) \(3C{H_3}COOH + Fe{(OH)_3}\xrightarrow{{}}{(C{H_3}COO)_3}Fe + 3{H_2}O\)

c) \(2{C_2}{H_5}OH + 2Na\xrightarrow{{}}2{C_2}{H_5}ONa + {H_2}\) 

d) \(6C{H_3}COOH + F{e_2}{O_3}\xrightarrow{{}}2{(C{H_3}COO)_3}Fe + 3{H_2}O\)

2)

Phản ứng xảy ra:

\(2C{H_3}COOH + Mg\xrightarrow{{}}{(C{H_3}COO)_2}Mg + {H_2}\)

Ta có: 

\({n_{Mg}} = \frac{{3,6}}{{24}} = 0,15{\text{ mol = }}{{\text{n}}_{{H_2}}}\xrightarrow{{}}{V_{{H_2}}} = 0,15.22,4 = 3,36{\text{ lít}}\)

Ta có: \({n_{C{H_3}COOH}} = 2{n_{Mg}} = 0,3{\text{ mol}} \to {{\text{C}}_{M{\text{ C}}{{\text{H}}_3}COOH}} = \frac{{0,3}}{{0,15}} = 2M\)

\(C{H_3}COOH + {C_2}{H_5}OH\xrightarrow{{}}C{H_3}COO{C_2}{H_5} + {H_2}O\)

Ta có: 

\({n_{C{H_3}COO{C_2}{H_5}}} = {n_{C{H_3}COOH}}.85\%  = 0,3.85\%  = 0,255{\text{ mol}} \to {{\text{m}}_{C{H_3}COO{C_2}{H_5}}} = 0,255.88 = 22,44{\text{ gam}}\)