$\dfrac{V_{C_2H_5OH}}{V_{rượu}} \times 100=90$

`=>`$V_{C_2H_5OH}=\dfrac{90 \times 90}{100}=81(L)=81000(mL)$

`=>`$m_{C_2H_5OH}=V \times D=81000 \times 0.8=64800(gam)$

`=>`$n_{C_2H_5OH}=\dfrac{32400}{23}(mol)$

$(C_6H_{10}O_5)_n \to nC_6H_{12}O_6 \to 2nC_2H_5OH+2nCO_2$

`=>`$n_{(C_6H_{10}O_5)_n}=\dfrac{1}{2n} \times \dfrac{32400}{23}(mol)$

Mà hiệu suất là $85\%:$

`=>`$n_{(C_6H_{10}O_5)_n}= \dfrac{3240000}{3910n}(mol)$

`->`$m_{(C_6H_{10}O_5)_n}= \dfrac{3240000}{3910n} \times 162n=\dfrac{524880000}{3910}(gam)$

$10\%$ tạp nữa nên $m_{tinh \,\, bột}=\dfrac{524880000}{3910} \times \dfrac{100}{90}$

$=\dfrac{583200000}{3910}(gam)$

$=\dfrac{583200}{3910}(kg)$

$≈149.156(kg)$