Question

In the first step in the industrial process for making nitric acid, ammonia reacts with oxygen in the presence of a suitable catalyst to form nitric oxide and water vapor.

_ NH₃ (g) + _ O₂ (g) → _ NO (g) + _ H₂O (g)

How many liters of NH₃ (g) at 850°C and 5.00 atm are required to react with 1.00 mol O₂ in this reaction?

Answer 1

Approximately 18.4 liters of NH₃ (g) at 850°C and 5.00 atm are required to react with 1.00 mol O₂ in the first step of the industrial process for making nitric acid.

Step-by-step explanation:

To determine the number of liters of NH₃ (g) required to react with 1.00 mol O₂, we need to use the balanced chemical equation and convert the moles of NH₃ to liters using the ideal gas law.

1. Balance the chemical equation: 4 NH₃ (g) + 5 O₂ (g) → 4 NO (g) + 6 H₂O (g)
2. Convert the moles of NH₃ to moles of O₂ using the stoichiometric ratio: 4 moles NH₃ = 5 moles O₂
3. Convert the moles of O₂ to liters using the ideal gas law: PV = nRT, where P = pressure, V = volume, n = number of moles, R = ideal gas constant, and T = temperature in Kelvin.

Let's assume the temperature is 850°C, which is 1123 K, and the pressure is 5.00 atm.

Using the ideal gas law equation, we can rearrange it to solve for volume (V): V = (nRT) / P

We know the number of moles of O₂ is 1.00 mol, R = 0.0821 L·atm/mol·K, and P = 5.00 atm. Substituting these values into the equation, we can calculate the volume of NH₃ (g) required.

V = (1.00 mol * 0.0821 L·atm/mol·K * 1123 K) / 5.00 atm = 18.4316 L

Therefore, approximately 18.4 liters of NH₃ (g) at 850°C and 5.00 atm are required to react with 1.00 mol O₂ in this reaction.