1)
Các phản ứng xảy ra:
\(2C{H_3}COOH + Mg\xrightarrow{{}}{(C{H_3}COO)_2}Mg + {H_2}\)
\(2C{H_3}COOH + {K_2}C{O_3}\xrightarrow{{}}2C{H_3}COOK + C{O_2} + {H_2}O\)
\(6C{H_3}COOH + F{{\text{e}}_2}{O_3}\xrightarrow{{}}2{(C{H_3}COO)_3}Fe + 3{H_2}O\)
\(C{H_3}COOH + {C_2}{H_5}OH\xrightarrow{{{H_2}S{O_4},{t^o}}}C{H_3}COO{C_2}{H_5} + {H_2}O\)
2)
Các phản ứng xảy ra:
\({(RCOO)_3}{C_3}{H_5} + 3{H_2}O\xrightarrow{{HCl}}3RCOOH + {C_3}{H_5}{(OH)_3}\)
\({(RCOO)_3}{C_3}{H_5} + 3NaOH\xrightarrow{{}}3RCOONa + {C_3}{H_5}{(OH)_3}\)
3)
Cái này phải 300 gam mới đúng
Phản ứng xảy ra:
\(2C{H_3}COOH + Mg\xrightarrow{{}}{(C{H_3}COO)_2}Mg + {H_2}\)
Ta có:
\({n_{Mg}} = \frac{{4,8}}{{24}} = 0,2{\text{ mol}}\)
\({m_{C{H_3}COOH}} = 300.12\% = 36{\text{ gam}}\)
\( \to {n_{C{H_3}COOH}} = \frac{{36}}{{60}} = 0,6{\text{ mol > 2}}{{\text{n}}_{Mg}}\)
Vậy \(CH_3COOH\) dư
\( \to {n_{C{H_3}COOH{\text{ dư}}}} = 0,6 - 0,2.2 = 0,2{\text{ mol}}\)
\( \to {m_{C{H_3}COOH{\text{ dư}}}} = 0,2.60 = 12{\text{ gam}}\)
Theo phản ứng
\({n_{{H_2}}} = {n_{{{(C{H_3}COO)}_2}Mg}} = {n_{Mg}} = 0,2{\text{ mol}}\)
\( \to {V_{{H_2}}} = 0,2.22,4 = 4,48{\text{ lít}}\)
\( \to {m_{{{(C{H_3}COO)}_2}Mg}} = 0,2.(59.2 + 24) = 28,4{\text{ gam}}\)
