Em tham khảo nha :

\(\begin{array}{l}
1)\\
a)\\
2{C_2}{H_5}OH + 2Na \to 2{C_2}{H_5}ONa + {H_2}\\
{n_{{C_2}{H_5}OH}} = \dfrac{{9,2}}{{46}} = 0,2mol\\
{n_{{H_2}}} = \dfrac{{{n_{{C_2}{H_5}OH}}}}{2} = 0,1mol\\
{V_{{H_2}}} = 0,1 \times 22,4 = 2,24l\\
b)\\
{C_2}{H_5}OH + {O_2} \to C{H_3}COOH + {H_2}O\\
{n_{C{H_3}COOH}} = {n_{{C_2}{H_5}OH}} = 0,2mol\\
{m_{C{H_3}COOH}} = 0,2 \times 60 = 12g\\
2)\\
a)\\
C{H_3}COOH + NaOH \to C{H_3}COONa + {H_2}O\\
{n_{C{H_3}COOH}} = \dfrac{6}{{60}} = 0,1mol\\
{n_{NaOH}} = {n_{C{H_3}COOH}} = 0,1mol\\
{V_{NaOH}} = \dfrac{{0,1}}{2} = 0,05l = 50ml\\
b)\\
2C{H_3}COOH + CaC{O_3} \to {(C{H_3}COO)_2}Ca + C{O_2} + {H_2}O\\
{n_{C{O_2}}} = \dfrac{{{n_{C{H_3}COOH}}}}{2} = 0,05mol\\
{V_{C{O_2}}} = 0,05 \times 24,79 = 1,2395l\\
3)\\
a)\\
{C_2}{H_5}OH + 3{O_2} \to 2C{O_2} + 3{H_2}O\\
{n_{{C_2}{H_5}OH}} = \dfrac{{9,2}}{{46}} = 0,2mol\\
{n_{{O_2}}} = 3{n_{{C_2}{H_5}OH}} = 0,6mol\\
{V_{{O_2}}} = 0,6 \times 22,4 = 13,44l\\
{V_{kk}} = 5{V_{{O_2}}} = 67,2l\\
b)\\
2{C_2}{H_5}OH + 2Na \to 2{C_2}{H_5}ONa + {H_2}\\
{n_{{H_2}}} = \dfrac{{{n_{{C_2}{H_5}OH}}}}{2} = 0,1mol\\
{V_{{H_2}}} = 0,1 \times 24,79 = 2,479l
\end{array}\)

Đáp án + Giải thích các bước giải:

Câu 1:

$a\big)$

$n_{C_2H_5OH}=\frac{9,2}{46}=0,2(mol)$

$2C_2H_5OH+2Na\to 2C_2H_5ONa+H_2$

Theo PT: $n_{H_2}=0,1(mol)$

$\to V_{H_2(đktc)}=0,1.22,4=2,24(l)$

$b\big)$

$C_2H_5OH+O_2\xrightarrow{men\,giấm}CH_3COOH+H_2O$

Theo PT: $n_{CH_3COOH}=0,2(mol)$

$\to m_{CH_3COOH}=0,2.60=12(g)$

Câu 2:

$a\big)$

$n_{CH_3COOH}=\frac{6}{60}=0,1(mol)$

$CH_3COOH+NaOH\to CH_3COONa+H_2O$

Theo PT: $n_{NaOH}=0,1(mol)$

$\to V_{dd\,NaOH}=\frac{0,1}{2}=0,05(l)=50(ml)$

$b\big)$

$CaCO_3+2CH_3COOH\to (CH_3COO)_2Ca+CO_2+H_2O$

Theo PT: $n_{CO_2}=0,05(mol)$

$\to V_{CO_2(đkc)}=0,05.24,79=1,2395(l)$

Câu 3:

$a\big)$

$n_{C_2H_5OH}=\frac{9,2}{46}=0,2(mol)$

$C_2H_5OH+3O_2\xrightarrow{t^o}2CO_2+3H_2O$

Theo PT: $n_{O_2}=3n_{C_2H_5OH}=0,6(mol)$

$\to V_{kk}=\frac{0,6.22,4}{20\%}=67,2(l)$

$b\big)$

$2C_2H_5OH+2Na\to 2C_2H_5ONa+H_2$

Theo PT: $n_{H_2}=0,1(mol)$

$\to V_{H_2(đkc)}=0,1.24,79=2,479(l)$