Question

An intermediate reaction used in the production of nitrogen-containing fertilizers is that between ammonia and oxygen: $$ 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) $$ A 150.0 - \(\mathrm{L}\) reaction chamber is charged with reactants to the following partial pressures at \(500^{\circ} \mathrm{C}: P_{\mathrm{NH}_{3}}=1.3 \mathrm{~atm}\) \(P_{\mathrm{O}_{2}}=1.5 \mathrm{~atm} .\) What is the limiting reactant?

Short answer

Answer: The limiting reactant in this reaction is oxygen (O₂).

Step-by-step solution

Write the balanced chemical equation

The balanced chemical equation for the reaction is already given: $$4 \mathrm{NH}_{3}(g) + 5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g) + 6 \mathrm{H}_{2} \mathrm{O}(g)$$

Determine the initial moles of NH₃ and O₂

We are given the initial partial pressures of NH₃ and O₂, as well as the volume of the reaction chamber, so we can use the Ideal Gas Law to find the initial moles of each reactant. Recall that the Ideal Gas Law is: $$PV = nRT$$ Where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is the temperature. Given that the temperature is \(500^{\circ} \mathrm{C}\), we need to convert this to Kelvin: $$T = 500 + 273.15 = 773.15\,\text{K}$$ Now we can find the initial moles of NH₃ (n(NH₃)) and O₂ (n(O₂)): $$n(\mathrm{NH}_3) = \dfrac{P_{\mathrm{NH}_3}V}{RT}$$ $$n(\mathrm{NH}_3) = \dfrac{1.3\,\text{atm}\cdot 150.0\,\text{L}}{0.0821\,\dfrac{\text{L}\cdot\text{atm}}{\text{mol}\cdot\text{K}}\cdot 773.15\,\text{K}}$$ $$n(\mathrm{NH}_3) = 3.61\,\text{mol}$$ $$n(\mathrm{O}_2) = \dfrac{P_{\mathrm{O}_2}V}{RT}$$ $$n(\mathrm{O}_2) = \dfrac{1.5\,\text{atm}\cdot 150.0\,\text{L}}{0.0821\,\dfrac{\text{L}\cdot\text{atm}}{\text{mol}\cdot\text{K}}\cdot 773.15\,\text{K}}$$ $$n(\mathrm{O}_2) = 4.14\,\text{mol}$$

Determine the required moles of each reactant

For every 4 moles of NH₃, we need 5 moles of O₂ for the reaction to proceed completely, as per the stoichiometry of the balanced equation. We can calculate the required moles of each reactant as follows: Required moles of NH₃ (n'(NH₃)): $$n'(\mathrm{NH}_3) = \dfrac{4}{5}\cdot n(\mathrm{O}_2)$$ $$n'(\mathrm{NH}_3) = \dfrac{4}{5}\cdot 4.14\,\text{mol}$$ $$n'(\mathrm{NH}_3) = 3.31\,\text{mol}$$ Required moles of O₂ (n'(O₂)): $$n'(\mathrm{O}_2) = \dfrac{5}{4}\cdot n(\mathrm{NH}_3)$$ $$n'(\mathrm{O}_2) = \dfrac{5}{4}\cdot 3.61\,\text{mol}$$ $$n'(\mathrm{O}_2) = 4.51\,\text{mol}$$

Identify the limiting reactant

Now we will compare the available moles of each reactant to the required moles based on the stoichiometric ratio to determine the limiting reactant: For NH₃: $$n(\mathrm{NH}_3) = 3.61\,\text{mol}$$ $$n'(\mathrm{NH}_3) = 3.31\,\text{mol}$$ Since we have more than enough moles of NH₃ available, NH₃ is not the limiting reactant. For O₂: $$n(\mathrm{O}_2) = 4.14\,\text{mol}$$ $$n’(\mathrm{O}_2) = 4.51\,\text{mol}$$ Since we do not have enough moles of O₂ available, O₂ is the limiting reactant. Thus, the limiting reactant in this reaction is oxygen (O₂).