Đáp án `+` Giải thích các bước giải:

Bài `1:`

`1. C_{12} H_{22} O_{11} + H_2 O` $\xrightarrow{axit, t^o}$ `C_6 H_{12} O_6 + C_6 H_{12} O_6`

`2. C_6 H_{12} O_6` $\xrightarrow[30-32^oC]{\text{men rượu}}$ `2C_2 H_5 OH + 2CO_2↑`

`3. C_2 H_5 OH + O_2` $\xrightarrow[25-30^oC]{\text{men giấm}}$ `CH_3 COOH + H_2 O`

`4. 2CH_3 COOH + Mg -> (CH_3 COO)_2 Mg + H_2↑`

`5. 2C_2 H_5 OH + 2K -> 2C_2 H_5 OK + H_2↑`

Bài `2:`

`n_{C_2 H_4} = (3,36)/(22,4) = 0,15 (mol)`

`a.`

`C_2 H_4 + Br_2 -> C_2 H_4 Br_2`

`b.`

Theo phương trình: `n_{C_2 H_4 Br_2} = n_{C_2 H_4} = 0,15 (mol)`

`-> m_{C_2 H_4 Br_2} = 0,15 . (12 . 2 + 1 . 4 + 80 . 2) = 28,2 (g)`

`c.`

Theo phương trình: `n_{Br_2} = n_{C_2 H_4} = 0,15 (mol)`

`-> V_{ddBr_2} = (0,15)/(0,1) = 1,5 (l)`

Đáp án + Giải thích các bước giải:

Bài 1:

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

$(2)C_6H_{12}O_6\xrightarrow[men\,rượu]{30-32^o}2C_2H_5OH+2CO_2$

$(3)C_2H_5OH+O_2\xrightarrow[men\,giấm]{25-30^0C}CH_3COOH+H_2O$

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

$(5)2K+2C_2H_5OH\to 2C_2H_5OK+H_2$

Bài 2:

$a\big)C_2H_4+Br_2\to C_2H_4Br_2$

$b\big)$

$n_{C_2H_4}=\frac{3,36}{22,4}=0,15(mol)$

Theo PT: $n_{C_2H_4Br_2}=n_{C_2H_4}=0,15(mol)$

$\to m_{C_2H_4Br_2}=0,15.188=28,2(g)$

$c\big)$

Theo PT: $n_{Br_2}=n_{C_2H_4}=0,15(mol)$

$\to V_{dd\,Br_2}=\frac{0,15}{0,1}=1,5(l)=1500(ml)$