Question

Answer 1

$V_{C_2H_5OH} = 460.\dfrac{50}{100} = 230(ml)$

$m_{C_2H_5OH} = D.V = 230.0,8 = 184(gam)$

$n_{C_2H_5OH} = \dfrac{184}{46} = 4(mol)$

$(C_6H_{10}O_5)_n + nH_2O \to nC_6H_{12}O_6$
$C_6H_{12}O_6 \xrightarrow{t^o,xt} 2CO_2 + 2C_2H_5OH$

$n_{tinh\ bột\ pư} = \dfrac{1}{2n}n_{C_2H_5OH} = \dfrac{2}{n}(mol)$

$n_{tinh\ bột\ đã\ dùng} = \dfrac{2}{n} : 50\% = \dfrac{4}{n}(mol)$
$m_{tinh\ bột} = \dfrac{4}{n}.162n = 648(gam)$

$m_{gạo\ nếp} = 648 : 80\% = 810(gam)$
Đáp án D

Answer 2

D. 810 gam