To solve this problem, we conduct stoichiometric calculations based on a balanced chemical equation. We find that 5,625 kg of nitrogen monoxide will be produced from the given reactants, and 1,312.5 kg of ammonia will remain as the excess reactant.
The problem involves a stoichiometric calculation to find the mass of nitrogen monoxide (NO) formed from the reaction of ammonia (NH₃) and oxygen (O₂), as well as determining the excess reagent, as per the following balanced chemical equation:
- 4NH₃ (g) + 5O₂ (g) → 4NO (g) + 6H₂O (l)
A mass of Nitrogen Monoxide Formed
To find the mass of NO formed, compare the molar mass and the stoichiometry of the reactants:
- Molar mass of NH₃ = 17 g/mol
- Molar mass of O₂ = 32 g/mol
Since 4500 kg of NH₃ is given (which is 4,500,000 g), we can calculate the number of moles of NH₃:
Moles of NH₃ = (4,500,000 g ÷ 17 g/mol) = 264,705.88 moles
Given 7500 kg of O₂ (which is 7,500,000 g), moles of O₂:
Moles of O₂ = (7,500,000 g ÷ 32 g/mol) = 234,375 moles
Using the stoichiometry of 4 moles of NH₃ to 5 moles of O₂, we find that O₂ is the limiting reagent:
Moles of NO produced by O₂ = (234,375 moles of O₂ ÷ 5) × 4 = 187,500 moles of NO
The mass of NO produced:
Mass of NO = (Moles of NO) × (Molar mass of NO)
Since the molar mass of NO is 30 g/mol:
Mass of NO = (187,500 moles) × 30 g/mol = 5,625,000 g or 5,625 kg
Excess Reagent Remaining
To find the mass of the excess reactant (NH₃), we need to subtract the moles of NH₃ that reacted based on the moles of the limiting reagent (O₂).
Moles of NH₃ reacted = (234,375 moles of O₂ ÷ 5) × 4 = 187,500 moles of NH₃
Mass of NH₃ reacted = (187,500 moles) × 17 g/mol = 3,187,500 g or 3,187.5 kg
Mass of NH₃ remaining = Initial mass of NH₃ - Mass of NH₃ reacted
Mass of NH₃ remaining = 4,500 kg - 3,187.5 kg = 1,312.5 kg