2)
\({C_{12}}{H_{22}}{O_{11}} + {H_2}O\xrightarrow{{axit}}{C_6}{H_{12}}{O_6} + {C_6}{H_{12}}{O_6}\)
\({C_6}{H_{12}}{O_6}\xrightarrow{{men}}2{C_2}{H_5}OH + 2C{O_2}\)
\({C_2}{H_5}OH + {O_2}\xrightarrow{{men}}C{H_3}COOH + {H_2}O\)
\(C{H_3}COOH + {C_2}{H_5}OH\xrightarrow{{{H_2}S{O_4},{t^o}}}C{H_3}COO{C_2}{H_5} + {H_2}O\)
3)
Phản ứng xảy ra:
\({C_2}{H_5}OH + {O_2}\xrightarrow{{men}}C{H_3}COOH + {H_2}O\)
\(C{H_3}COOH + KOH\xrightarrow{{}}C{H_3}COOK + {H_2}O\)
Ta có:
\({n_{KOH}} = {n_{C{H_3}COOH}} = {n_{C{H_3}COOK}} = {n_{{C_2}{H_5}OH}} = 0,9.0,5 = 0,45{\text{ mol}}\)
\({m_{{C_2}{H_5}OH}} = 0,45.46 = 20,7{\text{ gam}} \to {{\text{V}}_{{C_2}{H_5}OH}} = \frac{{20,7}}{{0,8}} = 25,875{\text{ ml}}\)
\( \to {V_A} = \frac{{0,45}}{{0,6}} = 0,75{\text{ lít}}\)
\({V_B} = {V_A} + {V_{KOH}} = 0,75 + 0,5 = 1,25{\text{ lít}} \to {{\text{C}}_{M{\text{C}}{{\text{H}}_3}COOK}} = \frac{{0,45}}{{1,25}} = 0,36M\)
