$a,PTPƯ:C_6H_{12}O_6\xrightarrow{men\ rượu,30-35^{o}C} 2C_2H_5OH+2CO_2↑$
$n_{CO_2}=\dfrac{6,72}{22,4}=0,3mol.$
$Theo$ $pt:$ $n_{C_2H_5OH}=n_{CO_2}=0,3mol.$
$⇒m_{C_2H_5OH}=0,3.46=13,8g.$
$b,Theo$ $pt:$ $n_{C_6H_{12}O_6}=\dfrac{1}{2}n_{CO_2}=0,15mol.$
Mà $H=80\%$ nên:
$⇒n_{C_6H_{12}O_6}=0,15.80\%=0,12mol.$
$⇒m_{C_6H_{12}O_6}=0,12.180=33,75g.$
chúc bạn học tốt!