Question

Urea \(\left(\mathrm{H}_{2} \mathrm{NCONH}_{2}\right)\) is used extensively as a nitrogen source in fertilizers. It is produced commercially from the reaction of ammonia and carbon dioxide: $$2 \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{NCONH}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(g)$$ Ammonia gas at \(223^{\circ} \mathrm{C}\) and \(90 .\) atm flows into a reactor at a rate of \(500 .\) L/min. Carbon dioxide at \(223^{\circ} \mathrm{C}\) and \(45\) atm flows into the reactor at a rate of \(600 .\) L/min. What mass of urea is produced per minute by this reaction assuming \(100 \%\) yield?

Short answer

First, we can find the moles of ammonia (NH₃) and carbon dioxide (CO₂) flowing into the reactor per minute using the ideal gas law: \(n = \frac{PV}{RT}\). For NH₃: \(n_{NH_3} = \frac{(90\,\text{atm})(500\,\text{L/min})}{(0.0821\,\text{L atm/mol K})(496\,\text{K})}\) and for CO₂: \(n_{CO_2} = \frac{(45\,\text{atm})(600\,\text{L/min})}{(0.0821\,\text{L atm/mol K})(496\,\text{K})}\). Next, determine the limiting reactant by comparing the stoichiometric ratio to the actual ratio. Then, calculate the moles of urea produced per minute using the limiting reactant and stoichiometric coefficients. Finally, find the mass of urea produced per minute by multiplying the moles of urea produced per minute by the molar mass of urea (60 g/mol).

Step-by-step solution

Convert flow rates into moles

To find the moles of chemicals flowing into the reactor per minute, we first need to convert the volume flow rate (L/min) into moles by using the ideal gas law, PV=nRT. For ammonia and carbon dioxide, the moles can be calculated as follows: For Ammonia (NH₃): $$PV = nRT \Rightarrow n = \frac{PV}{RT}$$ For Carbon Dioxide (CO₂): $$PV = nRT \Rightarrow n = \frac{PV}{RT}$$ Here, P is pressure, V is volume flow rate, R is the gas constant (0.0821 L atm/mol K), and T is the temperature in Kelvin (496K).

Determine the limiting reactant

Now that we have determined the moles of both reactants (NH₃ and CO₂) flowing into the reactor per minute, we can compare the molar ratio of the reactants to the stoichiometric ratio in the balanced equation. By doing this, we will be able to determine the limiting reactant, which will dictate the maximum amount of urea that can be produced. Stoichiometric ratio of NH₃ to CO₂: $$\frac{2 \text{ moles of NH}_{3}}{1 \text{ mole of CO}_{2}}$$ Actual ratio of NH₃ to CO₂: $$\frac{\text{moles of NH}_{3} \text{ flowing per minute}}{\text{moles of CO}_{2} \text{ flowing per minute}}$$ Compare the stoichiometric ratio to the actual ratio to determine which reactant is the limiting reactant.

Calculate the mass of urea produced per minute

Finally, using the limiting reactant and the stoichiometric coefficients from the balanced equation, we can calculate the moles of urea produced per minute. Moles of urea produced per minute: $$\frac{\text{moles of limiting reactant}}{\text{stoichiometric coefficient of limiting reactant}} \times \text{stoichiometric coefficient of urea}$$ To convert the moles of urea produced per minute to mass, we need to multiply the moles by the molar mass of urea: Mass of urea produced per minute: $$\text{moles of urea produced per minute} \times \text{molar mass of urea}$$ Molar mass of urea (H₂NCONH₂) = (2 x H) + (2 x N) + (1 x C) + (1 x O) = 60 g/mol By knowing the moles of urea produced per minute and the molar mass of urea, we can find the mass per minute.

Key concepts

Ideal Gas Law

Understanding the ideal gas law is crucial when dealing with gases, as it relates the pressure, volume, temperature, and number of moles of gas within a system. The law is typically represented by the equation, PV=nRT, where P stands for pressure, V represents the volume, n is the number of moles, R is the universal gas constant, and T indicates the temperature in Kelvin.

To apply this law to practical scenarios, such as in the urea production exercise, converting the flow rates of gases (ammonia and carbon dioxide) into moles is essential. This conversion allows for accurate calculation and comparison of reactants in a chemical process. It's a fundamental step to ensure the stoichiometry of the reaction is quantitatively correct. Remember, working in the correct units is critical; in this case, we use the temperature in Kelvin and pressure in atmospheres.

Limiting Reactant

In chemical reactions, the limiting reactant is the substance that is completely consumed first, thus dictating the maximum amount of product that can form. Identifying the limiting reactant requires a comparison between the molar amounts of each reactant and their stoichiometric ratios from the balanced chemical equation.

When analyzing the flow rates of ammonia and carbon dioxide, students must check against the stoichiometric requirement, which is a 2:1 ratio considering the balanced equation for urea synthesis. By comparing the actual mole ratio to the required stoichiometric ratio, it becomes clear which reactant limits the reaction. This step is pivotal as it determines how much urea can ultimately be produced, assuming 100% yield in the exercise scenario.

Stoichiometry

Stoichiometry is the section of chemistry that deals with the quantitative relationships, or ratios, within a chemical reaction. It's about converting moles of reactants to moles of products using the balanced chemical equation, providing a clear picture of how reactants are consumed and products are formed.

In applications like urea production, stoichiometry reveals how many moles of urea we can expect to form from the moles of the limiting reactant. The conversion process also utilizes the stoichiometric coefficients from the balanced equation, ensuring that the law of conservation of mass holds, and the reaction adheres to the prescribed molar ratios.

Molar Mass

The molar mass of a compound is the mass of one mole of that substance, typically expressed in grams per mole (g/mol). It's a fundamental concept for converting between moles of a substance and its mass, enabling chemists to precisely weigh out amounts of reactants and products.

In the context of the urea production problem, once the moles of urea are calculated using stoichiometry, knowing urea's molar mass (60 g/mol) allows for the conversion of moles to grams. This final calculation provides the mass of urea formed per minute under the given conditions, which is the desired outcome for the exercise. Molar mass is calculated by summing the masses of individual atoms (hydrogen, nitrogen, carbon, and oxygen) within the urea molecule, using their respective atomic weights from the periodic table.