Đáp án:
$a,m_{C_2H_5OH}=11,5g.$
$b,m_{C_6H_{12}O_6}=22,5g.$
Giải thích các bước giải:
$a,$
$Glucozơ:$ $C_6H_{12}O_6$
$PTPƯ:$
$C_6H_{12}O_6\xrightarrow{men\ rượu} 2CO_2+2C_2H_5OH$ $(1)$
$CO_2+Ba(OH)_2\xrightarrow{} BaCO_3↓+H_2O$ $(2)$
$n_{BaCO_3}=\dfrac{49,25}{197}=0,25mol.$
$Theo$ $pt2:$ $n_{CO_2}=n_{BaCO_3}=0,25mol.$
$Theo$ $pt1:$ $n_{C_2H_5OH}=n_{CO_2}=0,25mol.$
$⇒m_{C_2H_5OH}=0,25.46=11,5g.$
$b,Theo$ $pt1:$ $n_{C_6H_{12}O_6}=\dfrac{1}{2}n_{CO_2}=0,125mol.$
$⇒m_{C_6H_{12}O_6}=0,125.180=22,5g.$
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