GATE 2026 CH – Question 57
A feed containing 10 mol % propylene, 15 mol % ammonia and rest air is charged into a reactor to produce acrylonitrile (C$_3$H$_3$N) according to the following reaction.
$2C_3H_6 + 2NH_3 + 3O_2 \to 2C_3H_3N + 6H_2O$
Assume, air contains 80 mol % N$_2$ and 20 mol % O$_2$.
If the conversion of the limiting reactant is 30%, the composition of C$_3$H$_3$N (in mol %) in the product stream is ___________ (rounded off to two decimal places).
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Correct answer: 2.95 to 2.97
Explanation
Take 100 mol of feed: 10 mol C$_3$H$_6$, 15 mol NH$_3$ and 75 mol air (15 mol O$_2$ and 60 mol N$_2$). Dividing by the stoichiometric coefficients, C$_3$H$_6$ gives $10/2 = 5$, NH$_3$ gives $15/2 = 7.5$ and O$_2$ gives $15/3 = 5$. So the limiting reactant is C$_3$H$_6$ (O$_2$ equal to it). The extent of reaction is $\xi = 0.3 \times 5 = 1.5$. The product has $2\xi = 3$ mol C$_3$H$_3$N. The total moles change by $(2 + 6) - (2 + 2 + 3) = +1$ per unit extent, so the total is $101.5$ mol. The mole percent of C$_3$H$_3$N is $\frac{3}{101.5} \times 100 = 2.96$ %.