Tiến hành lên men m gam glucozơ thành C2H5OH với hiệu suất 75%

a) Hướng dẫn giải: 

$\Rightarrow n_{C_6{H}_{12}{O}_6}=\frac{m}{180}\left(mol\right)$

$\Rightarrow {\text{n}}_{{\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\text{\hspace{0.17em}\hspace{0.17em}phanung}}\text{=\hspace{0.17em}\hspace{0.17em}75\%.}\frac{\text{m}}{\text{180}}\text{\hspace{0.17em}\hspace{0.17em}=\hspace{0.17em}\hspace{0.17em}}\frac{m}{240}\text{(mol)}$

${\text{n}}_{\text{NaOH}}\text{ =\hspace{0.17em}\hspace{0.17em}0,5.2 = 1(mol)}$

$m_{\text{dd}NaOH}=D.V=2000.1,05=2100gam$

Gọi số mol CO₂ ở (2) và (3) lần lượt là x và y mol (x, y > 0)

$\begin{array}{l}{C}_6{H}_{12}{O}_6\stackrel{lenmen}{\to }2C_2{H}_{5}OH+2C{O}_{2 }\left(1\right)\\ \frac{m}{240}\to \frac{m}{120}\to \frac{m}{120}mol\end{array}$

$\begin{array}{l}C{O}_2+NaOH\to NaHC{O}_3 \left(2\right)\\ xx xmol\end{array}$

$\begin{array}{l}C{O}_2+2NaOH\to N{a}_{2}C{O}_3+H_{2}O \left(3\right)\\ y2yymol\end{array}$

Theo đề bài ta có: x + 2y = 1                       (I)

  $m_{ddsauphanung}=m_{C{O}_2}+m_{ddNaOH}=44(x+y)+2100\left(gam\right)$

$\begin{array}{l}\Rightarrow \frac{84x+106y}{44x+44y+2100}\times 100\%=3,211\%\\ \iff 82,58716x+104,58716y=67,431\left(II\right)\end{array}$

Từ (I) và (II) ta có $\left\{\begin{array}{l}x=0,5\\ y=0,25\end{array}\right.$

 $\begin{array}{l}{\text{n}}_{{\text{CO}}_{\text{2}}}\text{= x\hspace{0.17em}\hspace{0.17em}+\hspace{0.17em}\hspace{0.17em}y\hspace{0.17em}\hspace{0.17em}=\hspace{0.17em}\hspace{0.17em}0,5\hspace{0.17em}\hspace{0.17em}+\hspace{0.17em}\hspace{0.17em}0,25\hspace{0.17em}\hspace{0.17em}=\hspace{0.17em}\hspace{0.17em}0,75(mol)\hspace{0.17em}\hspace{0.17em}=\hspace{0.17em}\hspace{0.17em}}\frac{m}{120}\\ \Rightarrow m=90gam\end{array}$