Đáp án + Giải thích các bước giải:

Câu 4:

$(1)C_{12}H_{22}O_{11}+H_2O\xrightarrow{t^o,H^+}C_6H_{12}O_6+C_6H_{12}O_6$

$(2)C_6H_{12}O_6\xrightarrow[men\,rượu]{t^o}2C_2H_5OH+2CO_2$

$(3)C_2H_5OH+O_2\xrightarrow{men\,giấm}CH_3COOH+H_2O$

$(4)2CH_3COOH+Mg\to (CH_3COO)_2Mg+H_2$

Câu 5:

$(1)C_6H_{12}O_6\xrightarrow[men\,rượu]{t^o}2C_2H_5OH+2CO_2$

$(2)C_2H_5OH+O_2\xrightarrow{men\,giấm}CH_3COOH+H_2O$

$(3)CH_3COOH+C_2H_5OH\buildrel{{t^o,H_2SO_{4(đ)}}}\over\rightleftharpoons CH_3COOC_2H_5+H_2O$

$(4)CH_3COOC_2H_5+NaOH\xrightarrow{t^o}CH_3COONa+C_2H_5OH$

Câu 6:

$(1)CH_3COOC_2H_5O+NaOH\xrightarrow{t^o}CH_3COONa+C_2H_5OH$

$(2)2CH_3COONa+H_2SO_4\to 2CH_3COOH+Na_2SO_4$

$(3)2CH_3COOH+CuO\to  (CH_3COO)_2Cu+H_2O$

$(4)2C_2H_5OH+2Na\to 2C_2H_5ONa+H_2$

Đáp án+ Giải thích các bước giải:

`4)`

`C_{12}H_{22}O_{11} + H_2O` $\xrightarrow[Axit]{t^o}$ `C_6H_{12}O_6 + C_6H_{12}O_6`

`C_6H_{12}O_6` $\xrightarrow[30-35^oC]{\text{Men rượu}}$ `2C_2H_5OH + 2CO_2`

`C_2H_5OH + O_2` $\xrightarrow{\text{Men giấm}}$ `CH_3COOH + H_2O`

`2CH_3COOH + Mg -> (CH_3COO)_2Mg + H_2`

`5)`

`C_6H_{12}O_6` $\xrightarrow[30-35^oC]{\text{Men rượu}}$ `2C_2H_5OH + 2CO_2`

`C_2H_5OH + O_2` $\xrightarrow{\text{Men giấm}}$ `CH_3COOH + H_2O`

`CH_3COOH + C_2H_5OH` $\buildrel{{H_2SO_4(đ),t^o}}\over\rightleftharpoons$ `CH_3COOC_2H_5 + H_2O`

`CH_3COOC_2H_5 + NaOH` $\xrightarrow{t^o}$ `CH_3COONa + C_2H_5OH`

`6)`

`(1) CH_3COOC_2H_5 + NaOH` $\xrightarrow{t^o}$ `CH_3COONa + C_2H_5OH`

`(2) CH_33COONa + HCl -> CH_3COOH + NaCl`

`(3) 2CH_3COOH + CuO -> (CH_3COO)_2Cu + H_2O`

`(4) CH_3COOC_2H_5 + NaOH` $\xrightarrow{t^o}$  `C_2H_5OH + CH_3COONa `

`(5) 2C_2H_5OH + 2Na -> 2C_2H_5ONa + H_2`

$\text{#Cyrus}$