Question

Nitrogen monoxide is a pollutant commonly found in smokestack emissions. One way to remove it is to react it with ammonia. The balanced chemical equation for the reaction is 4NH₃(g) + 6NO(g) → 5N₂(g) + 6H₂O(ℓ). How many liters of ammonia are required to change 18.8 L of nitrogen monoxide to nitrogen gas?

Answer 1

To remove 18.8 liters of nitrogen monoxide, approximately 12.53 liters of ammonia are required, based on the stoichiometry of the balanced chemical reaction and using the proportion between the reactants.

Step-by-step explanation:

To calculate how many liters of ammonia are required to react with 18.8 liters of nitrogen monoxide, we can use the stoichiometry of the balanced chemical equation provided: 4NH₃(g) + 6NO(g) → 5N₂(g) + 6H₂O(ℓ). According to the equation, 4 moles of NH₃ react with 6 moles of NO. Since gases at the same temperature and pressure have the same volume per mole (by Avogadro's law), we can set up a proportion to solve for the volume of NH₃ needed.

The reaction ratio is 4:6 or 2:3 for NH₃:NO. Therefore, for every 3 liters of NO, 2 liters of NH₃ are required. If 18.8 liters of NO are involved, we can find the volume of NH₃ needed by setting up the proportion:

3 liters NO : 2 liters NH₃ = 18.8 liters NO : x liters NH₃

From this, x = (2/3) * 18.8 = 12.533... liters.

To remove 18.8 liters of nitrogen monoxide and convert it to nitrogen gas, you would need approximately 12.53 liters of ammonia.