`Na_2CO_3+2CH_3COOH→2CH_3COONa+H_2O+CO_2`

`a)`

`n_(Na_2CO_3)=(21,1)/106=0,2(mol)`

`V_(CO_2)=0,2.22,4=4,48(l)`

`b)`

`n_(CH_3COOH)=0,2.2=0,4(mol)`

`Vdd_(CH_3COOH)=(0,4)/1=0,4(l)`

`c)`

`C_2H_5OH+O_2→CH_3COOH+H_2O`

`m_(C_2H_5OH)=0,4.46=18,4(g)`

Đáp án:

\({V_{C{O_2}}} = 4,48{\text{ lít}}\)

\({V_{C{H_3}COOH}} = 0,4{\text{ lít}}\)

\({{\text{m}}_{{C_2}{H_5}OH}}  = 18,4{\text{ gam}}\)

Giải thích các bước giải:

Hình như là 21,2 gam chứ

 Phản ứng xảy ra:

\(N{a_2}C{O_3} + 2C{H_3}COOH\xrightarrow{{}}2C{H_3}COONa + C{O_2} + {H_2}O\)

\( \to {n_{N{a_2}C{O_3}}} = \frac{{21,2}}{{106}} = 0,2{\text{ mol = }}{{\text{n}}_{C{O_2}}}\)

\( \to {V_{C{O_2}}} = 0,2.22,4 = 4,48{\text{ lít}}\)

\( \to {n_{C{H_3}COOH}} = 2{n_{C{O_2}}} = 0,2.2 = 0,4{\text{ mol}}\)

\( \to {V_{C{H_3}COOH}} = \frac{{0,4}}{1} = 0,4{\text{ lít}}\)

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

\( \to {n_{{C_2}{H_5}OH}} = {n_{C{H_3}COOH}} = 0,4{\text{ mol}} \to {{\text{m}}_{{C_2}{H_5}OH}} = 0,4.46 = 18,4{\text{ gam}}\)