Đáp án:

$\rm a)$

Phương trình phản ứng:

$\rm C_6H_{12}O_6\xrightarrow{enzyme,\ t^°}2C_2H_5OH+2CO_2$

$\rm b)\\n_{CO_2}=\dfrac{3,36}{22,4}=0,15\ (mol)$

Theo phương trình:

$\rm n_{C_6H_{12}O_6}=\dfrac{1}{2}n_{CO_2}=0,075\ (mol)$

Khối lượng glucose cần dùng:

$\rm m_{C_6H_{12}O_6}=\dfrac{0,075×180}{60\%}=22,5\ (g)$

$\rm c)$

Theo phương trình:

$\rm n_{C_2H_5OH}=n_{CO_2}=0,15\ (mol)$

$\rm m_{C_2H_5OH}=0,15×46=6,9\ (g)$

$\rm V_{C_2H_5OH}=\dfrac{6,9}{0,8}=8,625\ (mL)\\V_{H_2O}=\dfrac{200}{1}=200\ (mL)$

$\rm G^°=\dfrac{8,625}{8,625+200}×100=4,13^°$

$\rm n_{CO_2}=\dfrac{3,36}{22,4}=0,15(mol)$ 

$\rm C_6H_{12}O_6 \xrightarrow{men-rượu} 2C_2H_5OH +2CO_2$ 

theo pthh 

$\rm n_{C_6H_{12}O_6 }=\dfrac{1}{2}n_{CO_2}=0,075(mol)$ 

$\rm n_{C_2H_5OH}=n_{CO_2}=0,15(mol)$ 

`b)` 

$\rm m_{C_6H_{12}O_6}=0,075.180=13,5(g)$ 

ta có $\rm H=60\%$ 

`->`$\rm m_{C_6H_{12}O_6 (tt)}=\dfrac{13,5}{60\%}=22,5(g)$ 

`c)` 

$\rm m_{C_2H_5OH}=0,15.46=6,9(g)$ 

`->`$\rm V_{C_2H_5OH}=\dfrac{6,9}{0,8}=8,625(ml)$

$\rm V_{H_2O}=\dfrac{200}{1}=200(ml)$ 

`->`$độ \ rượu =\dfrac{8,625}{8,625+200}.100=4,134^0$