Đáp án:

$\rm a)\\n_{CO_2}=\dfrac{17,92}{22,4}=0,8\ (mol)$

Phản ứng xảy ra:

$\rm C_6H_{12}O_6\xrightarrow{enzim,\ t^°}2CO_2+2C_2H_5OH$

Theo phản ứng:

$\rm n_{C_6H_{12}O_6}=\dfrac{1}{2}×n_{CO_2}=0,4\ (mol)$

$\rm n_{C_6H_{12}O_6\ (bd)}=\dfrac{0,4}{40\%}=1\ (mol)$

$\rm C_{M\ C_6H_{12}O_6}=\dfrac{1}{2}=0,5\ (M)$

$\rm b)\\V_{C_2H_5OH}=100×10^{3}×9\%=9000\ (mL)$

$\rm n_{C_2H_5OH}=\dfrac{9000×0,8}{46}=156,5\ (mol)$

Theo phản ứng:

$\rm n_{C_6H_{12}O_6}=\dfrac{1}{2}n_{C_2H_5OH}=78,25\ (mol)$

$\rm n_{C_6H_{12}O_6\ (bd)}=\dfrac{78,25}{90\%}=86,9\ (mol)$

$\rm m_{C_6H_{12}O_6}=86,9×180=15642\ (g)$