$a,PTPƯ:C_6H_{12}O_6\xrightarrow{men\ rượu} 2C_2H_5OH+2CO_2↑$
$b,n_{CO_2}=\dfrac{5,6}{22,4}=0,25mol.$
$Theo$ $pt:$ $n_{C_2H_5OH}=n_{CO_2}=0,25mol.$
$⇒m_{C_2H_5OH}=0,25.46=11,5g.$
$c,Theo$ $pt:$ $n_{C_6H_{12}O_6}=\dfrac{1}{2}n_{CO_2}=0,125mol.$
$⇒m_{C_6H_{12}O_6}=0,125.180=22,5g.$
Mà $H=85\%$ nên:
$⇒m_{C_6H_{12}O_6}=\dfrac{22,5}{85\%}=26,47g.$
chúc bạn học tốt!