Determination of Limiting Reactants in Ammonia and Oxygen Reactions

Understanding the concept of limiting reactants is crucial in chemical reactions, especially when dealing with gases like ammonia (NH3) and oxygen (O2). This article will explore how to determine the limiting reactant when 2 g of ammonia (NH3) is mixed with 4 g of oxygen (O2) under different reaction conditions and conditions.

Summary of Reactions

At room temperature, no reaction occurs between ammonia and oxygen, forming a mixture of ammonia and oxygen. However, when ammonia is burnt in pure oxygen, nitrogen gas (N2) and water vapor (H2O) are produced according to the reaction:

Reaction 1: 4NH3(g) 5O2(g) → 2N2(g) 6H2O(g)

When ammonia reacts with oxygen in the presence of a white-hot platinum catalyst, nitrogen monoxide (NO) gas and water vapor are formed as shown in:

Reaction 2: 4NH3(g) 5O2(g) → 4NO(g) 6H2O(g)

Calculation of Limiting Reactant

First, let's examine a simpler scenario: mixing 2 g of ammonia with 4 g of oxygen. To determine which of these substances is the limiting reagent, we need to calculate the number of moles of each reactant and compare them based on the balanced equation.

Molar mass of NH3 1 x 14.007 3 x 1.008 17.031 g/mol

Moles of Ammonia:

2 g of NH3 2/17.031 ≈ 0.117 mol

Moles of Oxygen:

4 g of O2 4/32 0.125 mol

Now, we will consider the stoichiometry of the reaction 4NH3(g) 5O2(g) → 4NO(g) 6H2O(g).

For every 4 moles of NH3, we need 5 moles of O2.

Number of moles of NH3 required 0.117 / 4 ≈ 0.03 mol

Number of moles of O2 required 0.125 / 5 0.025 mol

Since we have more moles of O2 than are needed to completely react with all the NH3, O2 is the excess reactant. Therefore, NH3 is the limiting reactant.

To find the excess remaining amount of ammonia:

Amount of NH3 remaining 0.117 - 0.03 0.087 mol or 1.485 g (mass 0.087 * 17.031)

Advanced Scenario

For a more complex scenario, consider 25 g of NH3 mixed with 4 mol of O2. In this case, we need to determine the limiting reagent using the concept of theoretical yield.

First, convert the mass of NH3 to moles:

25 g of NH3 25 g / 17.031 g/mol ≈ 1.468 mol

This is reacted with 4 mol of O2.

Using the balanced equation 4NH3(g) 5O2(g) → 4NO(g) 6H2O(g), the mole ratio of NH3 to NO is 4:4 or 1:1.

Theoretical yield of NO from NH3 1.468 mol.

Similarly, the molar ratio of O2 to NO is 5:4.

Theoretical yield of NO from O2 (4/5) * 4 mol 3.2 mol.

Since the reaction produces a smaller yield of NO from NH3, NH3 is the limiting reactant. The excess reactant is O2.

Conclusion

By understanding the stoichiometry and the limiting reactant concept, we can accurately predict the outcomes of chemical reactions involving ammonia and oxygen. This knowledge is essential for various industrial processes and scientific experiments.