$(1) (-C_6H_{10}O_5-)_n+nH_2O\xrightarrow{t^o,axit}nC_6H_{12}O_6$
$(2)C_6H_{12}O_6\xrightarrow{men rượu,30-35^oC}2C_2H_5OH+2CO_2$
`n_{C_2H_5OH}=920000:46=20000(mol)`
Qua 2 pt ta thấy:
`1(mol) (-C_6H_{10}O_5-)_n` tạo ra `2(mol)C_2H_5OH`
`⇒x(mol) (-C_6H_{10}O_5-)_n` tạo ra `20000(mol)C_2H_5OH`
`⇒x=10000(mol)`
$m_{(-C_6H_{10}O_5-)_n}=10000.162n=1620000n(g)=1620n(kg)$
Vì ngũ cốc (nc) chứa `81%` tinh bột nên:
`m_{nc}=1620n:81%=2000n(kg)`