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$a,PTPƯ:C_6H_{12}O_6\xrightarrow{lên\ men} 2C_2H_5OH+2CO_2↑$
$n_{CO_2}=\dfrac{11,2}{22,4}=0,5mol.$
$Theo$ $pt:$ $n_{C_2H_5OH}=n_{CO_2}=0,5mol.$
$⇒m_{C_2H_5OH}=0,5.46=23g.$
$b,Theo$ $pt:$ $n_{C_6H_{12}O_6}=\dfrac{1}{2}n_{CO_2}=0,25mol.$
$⇒m_{C_6H_{12}O_6}=0,25.276=69g.$
Mà $H=90\%$ nên:
$⇒m_{C_6H_{12}O_6}=\dfrac{69}{90\%}=76,67g.$
chúc bạn học tốt!