Đổi : $1.08(tấn)=1.08 \times 10^6(gam)$

$m_{xenlulozo}=50\%m_{\text{mùn cưa}}=1.08 \times 10^6 \times 50\%=540000(kg)$

$n_{xenlulozo}=\dfrac{m}{M}=\dfrac{m_{xenlulozo}}{162n}(mol)$

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

`->`$n_{C_2H_5OH}=2n \times \dfrac{m_{xenlulozo}}{162n} \times 65\%=\dfrac{13m_{xenlulozo}}{1620}(mol)$

`=>`$m_{C_2H_5OH}=nM=\dfrac{299m_{xenlulozo}}{810}(gam)$

$m_{C_2H_5OH}=V \times d$

`=>`$V=\dfrac{m_{C_2H_5OH}}{d}=\dfrac{\dfrac{299m_{xenlulozo}}{810}}{0.8}$

$=\dfrac{747500}{3}(mL)$

Mà $C_2H_5OH$ chiếm $5\%$ thể tích xăng sinh học $E_5$ hay:

$V_{C_2H_5OH}=5\%V_{\text{Xăng E_5}}$

`=>`$V_{\text{Xăng E_5}}=\dfrac{V_{C_2H_5OH}}{5\%}=\dfrac{\dfrac{747500}{3}}{0.05}$

$=\dfrac{747500}{0.15}(mL)$

$≈4983.3(L)$