Đáp án:

$\rm a)\ (\kern-6pt- C_6H_{10}O_5 -\kern-6pt)_n+nH_2O\xrightarrow{acid,\ t^°}nC_6H_{12}O_6\\b)\ C_6H_{12}O_6\xrightarrow[30-35^°C]{men}2C_2H_5OH+2CO_2$

Sơ đồ phản ứng:

$\rm (\kern-6pt- C_6H_{10}O_5-\kern-6pt)_n \xrightarrow{H\ =\ 80\%}nC_6H_{12}O_6\\\kern1cm1\kern4,3cm a\kern1cm(t)\\\kern,7cm162n\kern3,5cm180n\kern,6cm(t)$

$\rm a=\dfrac{180n}{162n}=\dfrac{10}{9}\ (t)$

$\rm m_{C_6H_{12}O_6\ (tt)}=\dfrac{10}{9}×80\%=\dfrac{8}{9}\ (t)$

Sơ đồ phản ứng:

$\rm C_6H_{12}O_6 \xrightarrow{H\ =\ 75\%}2C_2H_5OH\\\kern,7cm\dfrac{8}{9}\kern3,6cm b\kern,9cm(t)\\\kern,6cm180\kern3,3cm92\kern,809cm(t)$

$\rm b=\dfrac{92×\dfrac{9}{8}}{180}=\dfrac{184}{405}\ (t)$

$\rm m_{C_2H_5OH\ (tt)}=\dfrac{184}{405}×75\%=0,341\ (t)$