`m_{\text{tinh bột}}=15.85\%=12,75(g)`

PTHH:

$\rm (C_6H_{10}O_5)_n+nH_2O\xrightarrow{\text{H_2SO_4,t^o}} nC_6H_{12}O_6$

`n_{C_6H_{10}O_5}=(12,75)/(162)=(17)/(216)(mol)`

`=>n_{C_6H_{12}O_6(\text{lý thuyết})}=(17)/(216)(mol)`

Mà hiệu suất quá trình là `90\%`

`=>n_{C_6H_{12}O_6}=(17)/(216).90\%=(17)/(240)(mol)`

`=>m_{C_6H_{12}O_6}=(17)/(240).180=12,75(g)`