Đáp án:

\({C_{M{\text{ C}}{{\text{H}}_3}COOH}} = 1,6M\)

\(  {m_{{C_2}{H_5}OH}}  = 36,8{\text{ gam}}\)

\({m_{Zn}}  = 26{\text{ gam}}\)

\({m_{{{(C{H_3}COO)}_2}Zn}} = 73,2{\text{ gam}}\)

Độ rượu\(  = {15,33^o}\)

Giải thích các bước giải:

 Phản ứng xảy ra:

\({C_2}{H_5}OH + {O_2}\xrightarrow{{men}}C{H_3}COOH + {H_2}O\)

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

Ta có:

\({n_{{H_2}}} = \frac{{8,96}}{{22.4}} = 0,4{\text{ mol}}\)

\( \to {n_{C{H_3}COOH}} = 2{n_{{H_2}}} = 0,8{\text{ mol = }}{{\text{n}}_{{C_2}{H_5}OH}}\)

\({C_{M{\text{ C}}{{\text{H}}_3}COOH}} = \frac{{0,8}}{{0,5}} = 1,6M\)

\({n_{Zn}} = {n_{{{(C{H_3}COO)}_2}Zn}} = {n_{{H_2}}} = 0,4{\text{ mol}}\)

\( \to {m_{{C_2}{H_5}OH}} = 0,8.46 = 36,8{\text{ gam}}\)

\({m_{Zn}} = 0,4.65 = 26{\text{ gam}}\)

\({m_{{{(C{H_3}COO)}_2}Zn}} = 0,4.(59.2+65)=73,2{\text{ gam}}\)

\({V_{{C_2}{H_5}OH}} = \frac{{36,8}}{{0,8}} = 46{\text{ ml}}\)

Độ rượu\( = \frac{{46}}{{300}}.100 = {15,33^o}\)