Understanding Limiting Reactants in the Reaction Between NH? and O?
Understanding Limiting Reactants in the Reaction Between NH? and O?
When examining the reaction between ammonia (NH?) and oxygen (O?) as depicted in the equation 4NH? 5O? → 4NO 6H?O, it is essential to understand the concept of limiting reactants. This article explores the reaction where 8.00 moles of NH? and 10.00 moles of O? are used, and explains how to determine which reactant, if any, will be left unreacted.
Reaction Stoichiometry
The balanced chemical equation for the reaction is:
4NH? 5O? → 4NO 6H?O
Step-by-Step Analysis
Step 1: Calculate the Moles of O? Needed for NH?
According to the stoichiometric coefficients, the mole ratio of NH? to O? is 4:5. Therefore, for 8.0 moles of NH?, the amount of O? required can be calculated as:
Moles of O? required (5/4) x 8.0 moles 10.0 moles
Since the initial amount of O? given is exactly 10.0 moles, all of the O? will be used up in the reaction with NH?.
Step 2: Calculate the Moles of NH? Needed for O?
The mole ratio of O? to NH? is also 5:4. Thus, for 10.0 moles of O?, the amount of NH? required can be calculated as:
Moles of NH? required (4/5) x 10.0 moles 8.0 moles
Given that the initial amount of NH? is 8.0 moles, all of it will also be used up as it matches the required amount for the reaction.
Conclusion: Since the given amounts of NH? and O? are perfectly matched with the stoichiometric coefficients, neither reactant will be left over as an excess after the reaction. Both reactants will be completely consumed to form NO and H?O.
Molecular Mass and Product Calculation
For a more practical application, consider the mass of reactants and products. Ammonia (NH?) has a molar mass of approximately 17.01 g/mol, while nitrogen monoxide (NO) has a molar mass of approximately 30 g/mol.
If 17.01 g of NH? (1 mole) is oxidized, it would produce 30 g of NO. Following the stoichiometric ratio, the reaction of 8.0 moles of NH? would yield:
Moles of NO produced 8.0 moles
Mass of NO produced 8.0 moles x 30 g/mole 240 g
Similarly, the water (H?O) produced would be:
Moles of H?O produced 6 moles (as per balanced equation for 8.0 moles of NH?)
Mass of H?O produced 6 moles x 18 g/mole 108 g
This provides a clear understanding of how the reactants react completely in a 4:5 molar ratio to form the products.
Summary and Key Takeaways
Both NH? and O? will be fully consumed in the reaction when 8.00 moles of NH? and 10.00 moles of O? are used. The concept of limiting reactants is crucial in determining the limiting reagent and ensuring that no reactant remains unreacted, as seen in this stoichiometric analysis.
Related Keywords and References
limiting reactant NH? and O? reaction stoichiometric coefficientsRelated Articles:
Understanding Limiting Reactants in Scientific Reactions Exploring Stoichiometry in Chemical Reactions Introduction to Chemical Products Formed in Reactions