$(1) (-C_6H_{10}O_5-)_n+nH_2O\xrightarrow{t^o,axit}nC_6H_{12}O_6$

$(2)C_6H_{12}O_6\xrightarrow{men rượu,30-35^oC}2C_2H_5OH+2CO_2$

`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)`

1/

$-(C_{6}H_{10}O_{5})-_{n}+nH_{2}O \buildrel{{t^o,xt}}\over\longrightarrow nC_{6}H_{12}O_{6}\\ \text{10000/n                               10000}$ 

$C_{6}H_{12}O_{6} \buildrel{{\text{men rượu}}}\over\longrightarrow 2CO_{2} +2C_{2}H_{5}OH \\\text{10000                                  20000}$                                                   

2/

$nC_{2}H_{5}OH=\frac{920.1000}{46}=20000$

$m-(C_{6}H_{10}O_{5})-_{n}=\frac{10000}{n}.162n=1620000g=1620kg$

$m_{\text{ngũ cốc}}=\frac{1620}{81\%}=2000kg$