Answer:
Extent of reaction = 95.9 mol.
Fractional conversion of the limiting reactant = 0.846.
Percentage by which the other reactant is in excess = 25.2 %.
Explanation:
Hello.
In this case, for the undergoing chemical reaction:
[tex]CH_2=CH_2+HBr\rightarrow CH_3-CH_2Br[/tex]
We can write the mole balance per species also including the extent of the reaction:
[tex]CH_2=CH_2:A\\\\HBr: B\\\\CH_3-CH_2-Br:C[/tex]
[tex]x_AP=z_AF-\epsilon \\\\x_BP=z_BF-\epsilon \\\\x_CP=\epsilon[/tex]
Considering that P is the flow of the outlet product. In such a way, writing the data we know, we can write:
[tex]0.33P=z_A*265-\epsilon \\\\0.103P=z_B*265-\epsilon \\\\0.567P=\epsilon[/tex]
Whereas we can replace the C2H5Br mole balance in the others mole balances:
[tex]0.33P=z_A*265-0.567P \\\\0.103P=z_B*265-0.567P\\\\\\z_A*265-0.897P=0\\\\z_B*265-0.67P=0[/tex]
By knowing that [tex]z_B=1-z_A[/tex], we can write:
[tex]z_A*265-0.897P=0\\\\(1-z_A)*265-0.67P=0\\\\\\z_A*265-0.897P=0\\\\-z_A*265-0.67P=-265[/tex]
Thus, solving for P and [tex]z_A[/tex], we obtain:
[tex]z_A=0.572\\\\P=169.11mol[/tex]
It means that the extent of the reaction is:
[tex]\epsilon=0.567P=0.567*169.11mol\\\\\epsilon=95.9mol[/tex]
For the limiting reactant, due to the 1:1 mole ratio between the reactants, it is the one having the smallest flow rate:
[tex]F_A=0.572*265mol=151.58mol\\\\F_B=265mol-151.58mol=113.42mol[/tex]
It means that the limiting reactant is B which is HBr, whose fractional conversion is:
[tex]X_B=1-\frac{0.103*169.11}{113.42mol}\\ \\X_B=0.846[/tex]
Finally, the percentage by which the other reactant is in excess, corresponds to:
[tex]\% excess =(1-\frac{113.42mol}{151.58mol})*100\%\\ \\\%excess=25.2\%[/tex]
Regards.
