An important process for the production of acrylonitrile \(\left(\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{N}\right)\) is given by the following equation: $$2 \mathrm{C}_{3} \mathrm{H}_{6}(g)+2 \mathrm{NH}_{3}(g)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{C}_{3} \mathrm{H}_{3} \mathrm{N}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$$ A \(150 .\)-\(\mathrm{L}\)reactor is charged to the following partial pressures at \(25^{\circ} \mathrm{C}:\) $$\begin{aligned}P_{\mathrm{C}, \mathrm{H}_{6}} &=0.500 \mathrm{MPa} \\\P_{\mathrm{NH}_{3}} &=0.800 \mathrm{MPa} \\\P_{\mathrm{O}_{2}} &=1.500 \mathrm{MPa}\end{aligned}$$ What mass of acrylonitrile can be produced from this mixture \(\left(\mathrm{MPa}=10^{6} \mathrm{Pa}\right) ?\)

We will first convert the partial pressures of each reactant to the moles using the Ideal Gas Law equation: \[PV=nRT\] Where \(P\) is pressure, \(V\) is volume, \(n\) is the number of moles, \(R\) is the ideal gas constant (8.314 J/mol K), and \(T\) is the temperature. We are given the volume as 150 L, and the temperature as 25°C (which must be converted to Kelvin, 25 + 273.15 = 298.15 K). The ideal gas constant, R, will be in terms of L⋅Pa/mol⋅K; so, one must use the conversion 1000 L = 1 m³. Now, calculate the moles of each reactant. \[n_{C_3H_6} = \frac{0.500 \cdot 10^6 \cdot 0.150}{8.314 \cdot 298.15} \quad n_{NH_3} = \frac{0.800 \cdot 10^6 \cdot 0.150}{8.314 \cdot 298.15} \quad n_{O_2} = \frac{1.500 \cdot 10^6 \cdot 0.150}{8.314 \cdot 298.15}\]

Now, we will compare the stoichiometric ratios for each reactant to find which one is the limiting reactant. Using the balanced equation, the stoichiometric ratios are as follows: \[2 \mathrm{C}_{3} \mathrm{H}_{6}(g)+2 \mathrm{NH}_{3}(g)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{C}_{3} \mathrm{H}_{3} \mathrm{N}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\] Now, compare the mole ratios to their respective stoichiometric coefficients: \[\frac{n_{C_3H_6}}{2} \quad \frac{n_{NH_3}}{2} \quad \frac{n_{O_2}}{3}\] The reactant with the smallest result is the limiting reactant, so calculate the values using the moles calculated in step 1.

Using the limiting reactant and the stoichiometric ratio found by the balanced equation, we can calculate the number of moles of acrylonitrile (\(C_3H_3N\)) produced. Using the stoichiometric ratio, we can see that for example, for every 2 moles of propene, we get 2 moles of acrylonitrile produced.

Knowing the moles of acrylonitrile produced, we can calculate the mass by multiplying the moles by the molar mass of acrylonitrile, which is 53.06 g/mol: \[mass_{C_3H_3N} = moles_{C_3H_3N} \times 53.06\] Use the moles calculated in step 3 and apply the formula above to find the mass of acrylonitrile produced in grams. This is the final answer for the exercise.