62. B
$(C_6H_{10}O_5)_n+nH_2O\xrightarrow{enzim}nC_6H_12O_6$
$162n→18n→180n(mol)$
$C_6H_{12}O_6\xrightarrow{\text{men rượu}}2CO_2+2C_2H_5OH$
$180n→2.46n(mol)$
$⇒162n=2.46n$
$m_{\text{tinh bột}}=0,65.1=0,65(tấn)$
$m_{C_2H_5OH}=\dfrac{0,65.2.46n}{162n}.0,8≈0,29531(tấn)≈295,3(kg)$
$⇒B$
63. C
$m_{\text{tinh bột}}=(1-0,5).1=0,95(tấn)$
$(C_6H_{10}O_5)_n+nH_2O\xrightarrow{enzim}nC_6H_12O_6$
$162n→18n→180n(mol)$
$C_6H_{12}O_6\xrightarrow{\text{men rượu}}2CO_2+2C_2H_5OH$
$180n→2.46n(mol)$
$m_{C_2H_5OH}=\dfrac{0,95.2.46n}{162n}.0,85.0,85≈0,3898(tấn)≈389,8(kg)$
$⇒C$
64. B
$m_{\text{tinh bột}}=0,2.1=0,2(kg)$
$(C_6H_{10}O_5)_n+nH_2O\xrightarrow{enzim}nC_6H_12O_6$
$162n→18n→180n(mol)$
$m_{C_{6}H_{12}O_6}=\dfrac{0,2.180n}{162n}.0,81=0,18(kg)=180(g)$
$⇒B$
