$\rm n_{CO_2}=\dfrac{8,8}{44}=0,2(mol)$

 $\rm PTHH:C_6H_{12}O_6 \xrightarrow{men rượu,t^o} 2C_2H_5OH+2CO_2$

$\rm C_2H_5OH+O_2 \xrightarrow{men giấm,t^o} CH_3COOH+H_2O$
$\rm a.n_{C_2H_5OH}=n_{CO_2}=0,2(mol)$

$\rm m_{C_2H_5OH}=0,2.46=9,2(g)$

$\rm b.n_{C_6H_{12}O_6}=\dfrac{1}{2}n_{CO_2}=0,1(mol)$
$\rm H=70\%  \xrightarrow{} m_{C_6H_{12}O_6}=0,1.180.70\%=12,6(g)$

$\rm c.n_{CH_3COOH}=n_{C_2H_5OH}=0,2(mol)$

$\rm H=85\% \xrightarrow{} m_{CH_3COOH}=0,2.60.85\%=10,2(g)$

$\rm d.PTHH:C_6H_{12}O_6+Ag_2O \xrightarrow{NH_3} C_6H_{12}O_7+2Ag$

$\rm m_{C_6H_{12}O_7}=0,1.196.80\%=15,68(g)$

$\rm n_{Ag}=2n_{C_6H_{12}O_6}=0,2(mol)$

$\rm H=80\%=0,2.108.80\%=17,28(g)$

Đáp án:

Giải thích các bước giải:

 `a)`

Lên men glucozo : 

`C_6H_12O_6 overset(xt,t^o)(->) 2CO_2 + 2C_2H_5OH`(1)`

Lên men giấm : 

`C_2H_5OH + O_2 overset(xt,t^o)(->) CH_3COOH  + H_2O(2)`

`n_(CO_2) = 8.8/44 = 0.2 (mol)`

Theo PT `(1) : n_(CO_2) = n_(C_2H_5OH)  = 0.2 (mol)`

`=> m_(C_2H_5OH) = 0.2 * 46 = 9.2 (gam)`

`b)` 

Theo PT `(1) : 1/2 n_(CO_2) = n_(C_6H_12O_6) = 0.1 (mol)`

Thực tế : `n_(C_6H_12O_6) = 0.1 * 70% = 0.07 (mol)`

`=> m_(C_6H_12O_6) = 0.07 * 180 =12.6 (gam)`

`c)`

Theo PT `(2) : n_(C_2H_5OH) = n_(CH_3COOH) = 0.2 (mol)`

Thực tế : `n_(CH_3COOH) = 0.2 * 85% = 0.17 (mol)`

`=> m_(CH_3COOH) = 0.17 * 60 = 10.2 (gam)` 

`d)`

`C_6H_12O_6 + Ag_2O overset(xt)(->) C_6H_12O_7 + 2Ag ` 

Theo PT : `n_(C_6H_12O_6) = n_(C_6H_12O_7) = 0.1 (mol)`

`=> m_(C_6H_12O_7) = 0.1 * 196 * 80% = 15.68 (gam)`

Theo PT : `2n_(C_6H_12O_6) = n_(Ag) = 0.2 (mol)`

`=> m_(Ag) = 0.2 * 108 * 80% = 17.28(gam)`