Question

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 -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 \(\left(\mathrm{MPa}=10^{6} \mathrm{Pa}\right) ?\)

Short answer

The mass of acrylonitrile produced from the given mixture can be calculated using the limiting reactant and the stoichiometry of the balanced chemical equation. First, convert the initial partial pressures of the reactants to their moles. Then, determine the limiting reactant by comparing the mole ratios of the reactants. Use the moles of the limiting reactant to calculate the moles of acrylonitrile produced. Finally, find the mass of acrylonitrile by multiplying its moles by its molar mass (\(53.06 g/mol\)).

Step-by-step solution

Convert initial partial pressure to moles of each reactant

We will start by converting the given partial pressures to the number of moles for each reactant, using the ideal gas law: \(PV=nRT\). Rearranging for n, we get \(n=\frac{PV}{RT}\). Given total reaction pressure = \(2.8 MPa\), temperature = \(25°C\), and total volume = \(150 L\), we can find the moles of each reactant: - For propene, \(C_{3}H_{6}\): \(n_{C_{3}H_{6}}=\frac{0.5 \times 10^6 Pa \times 150 L}{8.314 J mol^{-1} K^{-1} (273.15 + 25) K}\) - For ammonia, \(NH_3\): \(n_{NH_3}=\frac{0.8 \times 10^6 Pa \times 150 L}{8.314 J mol^{-1} K^{-1} (273.15 + 25) K}\) - For oxygen, \(O_2\): \(n_{O_2}=\frac{1.5 \times 10^6 Pa \times 150 L}{8.314 J mol^{-1} K^{-1} (273.15 + 25) K}\)

Determine limiting reactant

To find the limiting reactant, we need to compare the mole ratio of each reactant from the balanced chemical equation. For the given balanced reaction, \(2 \mathrm{C}_{3} \mathrm{H}_{6} + 2 \mathrm{NH}_{3} + 3 \mathrm{O}_{2} \longrightarrow 2 \mathrm{C}_{3} \mathrm{H}_{3} \mathrm{N} + 6 \mathrm{H}_{2} \mathrm{O}\), we observe: - 2 moles of propene react with 2 moles of ammonia. - 2 moles of propene react with 3 moles of oxygen. - 3 moles of oxygen react with 2 moles of ammonia. We need to compare their given mole ratios with the calculated moles of reactants from Step 1. Whichever reactant has the smallest mole ratio will be the limiting reactant. Calculate and compare the mole ratios for each reactant pair: - Propene to Ammonia ratio: \(n_{C_3 H_6}:n_{NH_3}\) - Propene to Oxygen ratio: \(n_{C_3 H_6}:n_{O_2}\) - Oxygen to Ammonia ratio: \(n_{O_2}:n_{NH_3}\)

Calculate moles of acrylonitrile produced

Once we find the limiting reactant, we can use its moles to calculate the moles of acrylonitrile produced according to the stoichiometry of the balanced chemical equation. If X is the limiting reactant, then the moles of acrylonitrile produced can be calculated as follows: - If propene is the limiting reactant, \(n_{C_3 H_3 N} = n_{C_3 H_6}\). - If ammonia is the limiting reactant, \(n_{C_3 H_3 N} = n_{NH_3}\). - If oxygen is the limiting reactant, \(n_{C_3 H_3 N} = \frac{2}{3}n_{O_2}\).

Calculate the mass of acrylonitrile produced

Now that we have the moles of acrylonitrile produced, we can calculate the mass using the molar mass. For acrylonitrile (C3H3N), its molar mass is approximated as \(53.06 g/mol\). Therefore, the mass of acrylonitrile produced can be calculated as follows: \(m_{\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{N}} = n_{\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{N}} \times M_{\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{N}}\) where \(m_{\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{N}}\) is the mass of acrylonitrile, \(n_{\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{N}}\) is the calculated moles of acrylonitrile, and \(M_{\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{N}}\) is the molar mass of acrylonitrile. Obtain the final answer in grams for the mass of acrylonitrile produced from the given mixture.

Key concepts

Ideal Gas Law

The ideal gas law is a powerful equation of state that relates the pressure, volume, and temperature of an ideal gas to the number of moles of gas present. It is expressed as \(PV = nRT\), where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is the absolute temperature in Kelvin. This law allows us to calculate the number of moles of any gas when the other variables are known.

In the original exercise, the initial pressures of propene, ammonia, and oxygen were given. By applying the ideal gas law, we were able to determine how many moles of each of these gases were present in the reactor. The knowledge of moles is crucial, as it forms the basis for determining the extent of a chemical reaction by linking to stoichiometry and potential products. Remember, in practice, real gases may deviate slightly from the ideal gas behavior, especially at high pressures or low temperatures, but the ideal gas law remains a cornerstone in high-school to introductory college-level chemistry.

Limiting Reactant

In any chemical reaction, the limiting reactant is the substance that is completely consumed first, limiting the amount of product that can be formed. The limiting reactant determines the maximum possible yield of the reaction. To identify the limiting reactant, one must compare the mole ratios of the reactants from the balanced equation to the actual moles available of each reactant.

For the reaction in the exercise \(2 \mathrm{C}_{3} \mathrm{H}_{6} + 2 \mathrm{NH}_{3} + 3 \mathrm{O}_{2} \rightarrow 2 \mathrm{C}_{3} \mathrm{H}_{3} \mathrm{N} + 6 \mathrm{H}_{2} \mathrm{O}\), each reactant's mole ratio with respect to others can be determined. The reactant that provides the smallest stoichiometrically determined ratio is the limiting reactant. Once determined, the limiting reactant guides the calculation for the maximum amount of \(\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{N}\) that can be produced since it directly relates to the reaction's theoretical yield.

Molar Mass Calculation

Molar mass is a crucial concept in chemistry that allows you to relate the mass of a substance to the number of moles. Each element's atomic mass contributes to the molar mass of a compound, which can be found by summing the atomic masses of all atoms in a molecule. Molar mass is expressed in grams per mole (g/mol).

In the exercise, once the moles of acrylonitrile (\(\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{N}\)) are calculated using the limiting reactant, these moles are converted to mass using its molar mass of \(53.06\, \mathrm{g/mol}\). This conversion is done through the equation \(m = n \, \times \, M\), where \(m\) is the mass, \(n\) is the number of moles, and \(M\) is the molar mass. This step is usually the final one in stoichiometric calculations involving chemical reactions, allowing students to predict how much product will result under ideal conditions.