A,

PTHH:

$(C_6H_{10}O_5)_n+nH_2O\xrightarrow{enzyme}nC_6H_{12}O_6$

$C_6H_{12}O_6\xrightarrow{enzyme}2C_2H_5OH+2CO_2$

B,

$m_{\text{tinh bột}}=64\cdot75\%=48\ \text{kg}$

Hiệu suất quá trình: $H=100\%-19\%=81\%$

Sơ đồ:

$(C_6H_{10}O_5)_n\rightarrow nC_6H_{12}O_6\rightarrow 2nC_2H_5OH$

$162n\ \text{kg}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \rightarrow 92n\ \text{kg}$

$48\ \text{kg}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \rightarrow m_{lt}\ \text{kg}$

Khối lượng $C_2H_5OH$ lý thuyết:

$m_{lt}=\dfrac{48\cdot92n}{162n}=\dfrac{736}{27}\ \text{kg}$

Khối lượng $C_2H_5OH$ thực tế:

$m_{tt}=\dfrac{736}{27}\cdot81\%=22,08\ \text{kg}$

Thể tích $C_2H_5OH$ nguyên chất ($D=0,789\ \text{g/mL}=0,789\ \text{kg/L}$):

$V_{nc}=\dfrac{m_{tt}}{D}=\dfrac{22,08}{0,789}\ \text{L}$

Thể tích ethyl alcohol $46^\circ$:

$V_{46^\circ}=V_{nc}\cdot\dfrac{100}{46}=\dfrac{22,08}{0,789}\cdot\dfrac{100}{46}\approx60,84\ \text{L}$

Vậy thể tích ethyl alcohol $46^\circ$ thu được là $60,84\ \text{L}$.