Giải chi tiết câu 3 ,4,5 chi tiết giup mình

Gọi `X` là `RCOOR'`

`n_X=n_{KOH}=n_{ancol}=0, 1(mol) `

`=> R'+17=(4, 6) /(0, 1) => R'=29(C_2H_5)` 

`=> R+44+29=(8, 8) /(0, 1) => R=15(CH_3) `

`=> X` là `CH_3COOC_2H_5`

Câu 4:

`n_C=n_{CO_2}=(3, 36) /(22, 4) =0, 15(mol) ;n_H=2n_{H_2O}=2.(2, 7) /18=0,3(mol) `

`=> n_O=(3, 7-0, 15.12-0, 3) /16=0, 1(mol)` 

`=> n_C:n_H:n_O=0, 15:0, 3:0, 1=3:6:2`

`=> X` là `C_3H_6O_2`

Câu 5:

`C_6H_{12}O_6->2C_2H_5OH`

`n_{glu}=(2, 5.1000.20%) /180=25/9(mol) `

$\Rightarrow n_{C_2H_5OH(lt)}$`=50/9(mol) => n_{C_2H_5OH(tt) }=(50)/(9). 90%=5(mol) `

`=> m_{C_2H_5OH}=46.5=230g`

`=> V_{C_2H_5OH}=230/(0, 8) =287, 5(ml) `

`V_{dd}=(287,5.100)/40=718 ,75(ml) `

`3)`

`X:RCOOR'`

`->Y:R'OH`

`n_{KOH}=0,1.1=0,1(mol)`

`->n_X=n_Y=n_{KOH}=0,1(mol)`

`->M_Y=M_{R'}+17={4,6}/{0,1}=46`

`->M_{R'}=29(-C_2H_5)`

`->X:RCOOC_2H_5`

`->M_X=M_R+73={8,8}/{0,1}=88`

`->M_R=15(-CH_3)`

`->CH_3COOC_2H_5` (etyl axetat)

`4)`

`n_{CO_2}=n_{H_2O}=0,15(mol)`

`X:C_nH_{2n}O_2`

`->n_X={n_{CO_2}}/n={0,15}/n(mol)`

`->M_X=14n+32={3,7}/{{0,15}/n}={74}/{3}n`

`->n=3`

`->X:C_3H_6O_2`

`5)`

`n_{C_6H_{12}O_6}={2,5.1000.80\%}/{180}={100}/9(mol)`

`C_6H_{12}O_6`  $\xrightarrow{enzyme}$  `2C_2H_5OH+2CO_2`

`->n_{C_2H_5OH\ l t}=2n_{C_6H_{12}O_6}={200}/9(mol)`

`->V_{C_2H_5OH\ t t}={{200}/{9}.46.90\%}/{0,8}=1150(mL)`

`->V_r={1150.100}/{40}=2875(mL)`