Câu 1: 6nCO2 + 5nH2O > 6nO2 + ( C6H10O5 )n

          ( C6H10O5 )n + nH2O > n( C6H12O6 )

           C6H12O6 > 2C2H5OH + 2CO2

          2C2H5OH + 2Na > H2 + 2C2H5ONa

Câu 2 :

a)

 2C2H5OH+N2a--> 2C2H5ONa + H2

   x                                                 1/2x        mol

2CH3COOH+ 2Na ----> 2CH3COONa+H2

   y                                                        1/2 y mol

ta đc (46x+60y=15.2)

          (0.5x+0.5y=0.15)

=>x=0.2 y=0.1

%mC2H5OH=(46x0.2)/15.2x100=60.5%

=> %mCH3COOH= 100-60.5=39.5%

1)

Các phản ứng xảy ra:

\(6nC{O_2} + 5n{H_2}O\xrightarrow{{as}}{({C_6}{H_{10}}{O_5})_n}\)

\({({C_6}{H_{10}}{O_5})_n} + n{H_2}O\xrightarrow{{axit}}n{C_6}{H_{12}}{O_6}\)

\({C_6}{H_{12}}{O_6}\xrightarrow{{men}}2{C_2}{H_5}OH + 2C{O_2}\)

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

\(2{C_6}{H_{12}}{O_6} + {O_2}\xrightarrow{{xt}}2{C_6}{H_{12}}{O_7}\) (axit gluconic)

2)

Phản ứng xảy ra:

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

\(2C{H_3}COOH + 2Na\xrightarrow{{}}2C{H_3}COONa + {H_2}\)

Gọi số mol \(C_2H_5OH;CH_3COOH\) lần lượt là \(x;y\)

\( \to 46x + 60y = 15,2\)

Ta có:

\({n_{{H_2}}} = \frac{1}{2}x + \frac{1}{2}y = \frac{{3,36}}{{22,4}} = 0,15{\text{ mol}}\)

Giải được: \(x=0,2;y=0,1\)

\( \to {m_{{C_2}{H_5}OH}} = 0,2.46 = 9,2{\text{ gam}}\)

\( \to \% {m_{{C_2}{H_5}OH}} = \frac{{9,2}}{{15,2}} = 60,53\%  \to \% {m_{C{H_3}COOH}} = 39,47\% \)

Este hóa

\(C{H_3}COOH + {C_2}{H_5}OH\xrightarrow{{{H_2}S{O_4},{t^o}}}C{H_3}COO{C_2}{H_5} + {H_2}O\)

Vì \({n_{{C_2}{H_5}OH}} > {n_{C{H_3}COOH}}\) nên ancol dư, hiệu suất tính theo axit

\({n_{C{H_3}COO{C_2}{H_5}{\text{ lt}}}} = {n_{C{H_3}COOH}} = 0,1{\text{ mol}}\)

\( \to {n_{C{H_3}COO{C_2}{H_5}}} = 0,1.80\%  = 0,08{\text{ mol}}\)

\( \to {m_{C{H_3}COO{C_2}{H_5}}} = 0,08.88 = 7,04{\text{ gam}}\)