Final answer:
To find the mass of the excess reagent, we need to determine the limiting reagent first. In this case, ammonia is the limiting reagent, so there is no excess ammonia remaining. The mass of the excess oxygen is 4452.88 g.
Step-by-step explanation:
In order to determine the mass of the excess reagent, we first need to find the limiting reagent, which is the reactant that will be completely consumed in the reaction. To do this, we can calculate the amount of nitrogen monoxide and water that would be produced if both reactants were completely consumed.
First, we need to convert the masses of ammonia and oxygen to moles using their respective molar masses. Ammonia has a molar mass of 17 g/mol, so 4500 kg of ammonia is equal to 4500/17 = 264.71 mol. Oxygen has a molar mass of 32 g/mol, so 7500 kg of oxygen is equal to 7500/32 = 234.375 mol.
The balanced equation for the reaction is:
4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(l)
According to the balanced equation, the molar ratio between ammonia and nitrogen monoxide is 4:4, which simplifies to 1:1. Therefore, if 264.71 mol of ammonia reacts, 264.71 mol of nitrogen monoxide would be produced. Similarly, the molar ratio between ammonia and water is 4:6, which simplifies to 2:3. Therefore, if 264.71 mol of ammonia reacts, 397.06 mol of water would be produced.
Since the molar ratio between ammonia and nitrogen monoxide is 1:1, and the molar ratio between ammonia and water is 2:3, it means that ammonia is the limiting reagent in this reaction. This means that all of the ammonia will be completely consumed, and there will be no excess ammonia remaining.
Therefore, the mass of the excess reagent, which in this case is the oxygen, will be equal to the initial mass of oxygen minus the mass of oxygen that reacted with the ammonia.
The initial mass of oxygen is 7500 kg, which is equal to 7500/32 = 234.375 mol. Since the molar ratio between ammonia and oxygen is 4:5, we can use the given amount of ammonia (264.71 mol) to calculate the amount of oxygen that reacted. Therefore, 264.71 mol of ammonia would react with (264.71 mol * 5/4) = 330.88 mol of oxygen.
Now we can calculate the mass of the excess oxygen by subtracting the mass of oxygen that reacted with ammonia from the initial mass of oxygen. The molar mass of oxygen is 32 g/mol, so the mass of the oxygen that reacted is (330.88 mol * 32 g/mol) = 10,588.16 g. Therefore, the mass of the excess oxygen is equal to (234.375 mol * 32 g/mol) - 10,588.16 g = 4452.88 g.