$n_{CO_2}=\dfrac{44,8}{22,4}=2(mol)$ 

$\rm PTHH: C_6H_{12}O_6 \xrightarrow{to,xt} 2C_2H_5OH +2CO_2$ 

theo pthh 

$n_{C_2H_5OH} =n_{CO_2}=2(mol)$ 

`->`$m_{C_2H_5OH }=2.46 =92(g)$ 

b)

theo pthh 

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

`->`$m_{C_6H_{12}O_6}=1. 180=180(g)$ 

`=>`$m_{C_6H_{12}O_6(tt)}=\dfrac{180}{80\%}=225(g)$ 

$\rm n_{CO_2}=\dfrac{44,8}{22,4}=2(mol)$

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

$\rm a.n_{C_2H_5OH}=n_{CO_2}=2(mol)$

$\rm m_{C_2H_5OH}=2.46=92(g)$

$\rm b.n_{C_6H_{12}O_6} =\dfrac{1}{2} n_{CO_2}=1(mol)$

`=>`$\rm m_{C_6H_{12}O_6}lt=180.1=180(g)$

$\rm H=80\% \to m_{C_6H_{12}O_6}=180:80\%=225(g)$