Answer:
Explanation:
From the given information:
The two reactions occurring in the reactor are:
[tex]\mathtt{CH_3COOH + 2H_2 \to C_2H_5OH + H_2O}[/tex]
[tex]\mathtt{2C_2H_5OH \to (C_2H_5)_2O+H_2O}[/tex]
Let molar flowrate of [tex]\mathtt{CH_3COOH}[/tex] entering reactor = [tex]\mathtt{x_n}[/tex]
From the first reaction:
[tex]\mathtt{CH_3COOH + 2H_2 \to C_2H_5OH + H_2O}[/tex]
Since there is a 71.6% conversion of [tex]\mathtt{CH_3COOH}[/tex];
Then, amount of [tex]\mathtt{CH_3COOH}[/tex] remaining unreacted and exiting the reactor is:
= x - 0.716x
= 0.284x
Recall that; the component flow rate exiting the reactor = 3.81 mols
∴
0.284x = 3.81
x = 3.81/0.284
x = 13.42 moles
Thus, the molar flowrate of [tex]\mathtt{CH_3COOH}[/tex] entering the reactor is 13.42 moles