$(C_6H_10O_5)_n+H_2O\xrightarrow{\text{lên men}}C_6H_12O_6$
$C_6H_12O_6\xrightarrow{\text{men rượu}}C_2H_5OH+2CO_2$
-> $n_{(C_6H_10O_5)_n}=2n_{C_2H_5OH}$
$m_{\text{tinh bột}}=1.0,7=0,7(tấn)$
$n_{\text{tinh bột}}=\dfrac{0,7}{162n}=\dfrac{7}{1620n}(mol)$
$n_{C_2H_5OH}=2.\dfrac{7}{1620n}=\dfrac{14}{1620n}(mol)$
$m_{C_2H_5OH}\text{thực tế}=\dfrac{14}{1620n}.0,85.46≈0,338(tấn)$
=> Đáp án: A.0,338 tan