$(C_6H_{10}O_5)_n + nH_2O \xrightarrow{t^o} nC_6H_{12}O_6$
$C_6H_{12}O_6 \xrightarrow{t^o,xt} 2CO_2 + 2C_2H_5OH$
$m_{tinh\ bột} = 1000(100\% - 20\%) = 800(kg)$
$n_{tinh\ bột} = \dfrac{800}{162n}(kmol)$
$n_{glucozo} = n.n_{tinh\ bột} = \dfrac{800}{162}(kmol)$
$n_{C_2H_5OH} = 2n_{glucozo} = \dfrac{800}{81}(kmol)$
$n_{C_2H_5OH\ thu được} = \dfrac{800}{81} .70\% = \dfrac{560}{81}(kmol)$
$m_{C_2H_5OH} = \dfrac{560}{81}.46 = 318,02(kg)$
$V_{C_2H_5OH} = \dfrac{318,02}{0,8} = 397,25(lít)$
$V_{C_2H_5OH\ 60^o} = \dfrac{397,25.100}{60} = 662,55(lít)$
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