`a)`

Bình 1:

`m_{\text{bình tăng}}=m_{H_2O}=0,54(g)`

`->n_{H_2O}={0,54}/{18}=0,03(mol)`

Bảo toàn H: `n_H=2n_{H_2O}=0,06(mol)`

Bình 2:

`m_{\text{bình tăng}}=m_{CO_2}=0,88(g)`

`->n_{CO_2}={0,88}/{44}=0,02(mol)`

Bảo toàn C: `n_C=n_{CO_2}=0,02(mol)`

`->n_O={0,46-0,02.12-0,06}/{16}=0,01(mol)`

Đặt CTPT A là `C_xH_yO_z`

`->x:y:z=0,02:0,06:0,01=2:6:1`

`->` CTN: `(C_2H_6O)_n`

Mà `M_A=23.2=46(g//mol)`

`->(12.2+6+16).n=46`

`->n=1`

`->` CTPT A là `C_2H_6O`

`b)`

`A+Na->H_2↑`

`->A:\ CH_3-CH_2-OH`

`2CH_3-CH_2-OH+2Na->2CH_3-CH_2-ONa+H_2`

`c)`

`C_2H_5OH+CuO`  $\xrightarrow{t^o}$  `CH_3CHO+Cu+H_2O`

`CH_3CHO+2AgNO_3+3NH_3+H_2O->CH_3COONH_4+2NH_4NO_3+2Ag`

`d)`

`n_{C_2H_5OH\ l t}={9,2}/{46}=0,2(mol)`

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

Theo PT: `n_{CH_3COOH\ l t}=n_{C_2H_5OH\ l t}=0,2(mol)`

`->n_{CH_3COOH\ t t}=0,2.80\%=0,16(mol)`

`->m_{CH_3COOH\ t t}=0,16.60=9,6(g)`

`n_{C_2H_5OH\ du}=0,2.(100\%-80\%)=0,04(mol)`

`2CH_3COOH+2Na->2CH_3COONa+H_2`

`2C_2H_5OH+2Na->2C_2H_5ONa+H_2`

Theo PT: `\sum n_{H_2}=1/2n_{CH_3COOH\ t t}+1/2n_{C_2H_5OH\ du}=0,1(mol)`

`->V_{H_2}=0,1.22,4=2,24(L)`