$(C6H10O5)n$ + n$H2O$ => n$C6H12O6$
$C6H12O6$ => 2$C2H5OH$ +2$CO2$
n$(C6H10O5)n$=150/162n mol
=>n$C6H12O6$=150/162 mol
=>n$C2H5OH$=2n$C6H12O6$=1,852 mol
H%=81%=>n$C2H5OH$ thực tế=1,5 mol
=>m$C2H5OH$=1,5.46=69g
=>V$C2H5OH$ng chất=69/0,8=86,25ml
=>V$C2H5OH$ =86,25/46%=187,5ml