Question

In Ostwald's process for the manufacture of nitric acid, the first step involves the oxidation of ammonia gas by oxygen gas to give nitric oxide gas and steam. What is the maximum weight of nitric oxide that can be obtained starting only with \(10.00 \mathrm{~g}\). of ammonia and \(20.00 \mathrm{~g}\) of oxygen ?

Short answer

The maximum weight of nitric oxide obtained is 15.005 g.

Step-by-step solution

Write the Balanced Chemical Equation

The reaction is the oxidation of ammonia (H_3) by oxygen (O_2) to produce nitric oxide (NO) and water (H_2O). The balanced chemical equation is: \[4 \text{NH}_3 + 5 \text{O}_2 \rightarrow 4 \text{NO} + 6 \text{H}_2\text{O}\]

Calculate Molar Masses

Determine the molar masses of the substances involved: \( \text{NH}_3 \) has a molar mass of approximately 17.03 g/mol, \( \text{O}_2 \) has a molar mass of approximately 32.00 g/mol, and \( \text{NO} \) has a molar mass of approximately 30.01 g/mol.

Convert Masses to Moles

Calculate the number of moles of each reactant:\[\text{Moles of NH}_3 = \frac{10.00\, \text{g}}{17.03\, \text{g/mol}} \approx 0.587 \text{ mol}\] \[\text{Moles of O}_2 = \frac{20.00\, \text{g}}{32.00\, \text{g/mol}} = 0.625 \text{ mol}\]

Determine the Limiting Reactant

From the balanced equation, 4 moles of \( \text{NH}_3 \) react with 5 moles of \( \text{O}_2 \). Compare the mole ratio:- For \( \text{NH}_3 \), we need \( \frac{5}{4} \times 0.587 \approx 0.734 \text{ mol of } \text{O}_2\). - \( \text{O}_2 \) available is 0.625 mol.Oxygen is the limiting reactant because we need more oxygen than available.

Calculate Maximum Possible Moles of NO Produced

Using the limiting reactant \( \text{O}_2 \), use the mole ratio to find moles of \( \text{NO} \) produced:\[ \text{From}\, 5 \text{ moles of } \text{O}_2, \text{ we get } 4 \text{ moles of } \text{NO}\]So from 0.625 moles of \( \text{O}_2 \), the moles of \( \text{NO} \) is:\[ \frac{4}{5} \times 0.625 = 0.500 \text{ mol of } \text{NO}\]

Convert Moles of NO to Grams

Calculate the mass of \( \text{NO} \) produced using its molar mass:\[ \text{Mass of NO} = 0.500 \times 30.01\, \text{g/mol} = 15.005\, \text{g} \]

Key concepts

Limiting Reactant

To understand which reactant limits the reaction, it's essential to know the role of the limiting reactant. In any chemical reaction, the limiting reactant is the substance that is completely consumed first and thus determines the amount of product formed. This concept is crucial because it dictates the reaction's output regardless of the other reactants' quantities.
In the exercise from Ostwald's process, we identify the limiting reactant by comparing the mole ratio of the reactants based on the balanced chemical equation. For ammonia (NH₃) and oxygen (O₂), the balanced equation tells us that 4 moles of NH₃ require 5 moles of O₂.

Here's how it works:

• Calculate the moles of each reactant.
• Using the balanced equation, determine how many moles of the other reactant are needed to completely react with the amounts available.
• The reactant that provides a lower mole ratio (compared to what's required) is the limiting reactant.

In this problem, oxygen was found to be the limiting reactant, as it offers fewer moles than needed to react with all the available ammonia, ensuring that it dictates the reaction's completion.

Balanced Chemical Equation

A balanced chemical equation represents the conservation of atoms in a chemical reaction. It shows the precise relationship between reactants and products in terms of their chemical formulas and quantities. This equation is vital for calculations relating to limiting reactants and product amounts.
The balanced equation for the oxidation of ammonia by oxygen in Ostwald's process is:
\[4 \text{NH}_3 + 5 \text{O}_2 \rightarrow 4 \text{NO} + 6 \text{H}_2\text{O}\]

Here’s why balancing is crucial:

• It respects the Law of Conservation of Mass, ensuring that each type of atom that enters the reaction leaves it.
• Provides the mole ratio needed to convert moles of one substance to moles of another in stoichiometric calculations.
• Ensures that, if followed, calculations of reactants and products remain consistent and accurate.

By initially balancing the equation, it lays a foundational understanding for further stoichiometric calculations, such as determining the limiting reactant and the amount of product produced.

Molar Mass Calculation

Molar mass is a fundamental concept for converting a substance’s mass into moles, which is essential for stoichiometric calculations in chemistry. It is the mass of one mole of a chemical substance and is expressed in grams per mole. For effective calculations, knowing the molar mass of each substance involved in the reaction is crucial.

In this reaction:

• The molar mass of ammonia (NH₃) is about 17.03 g/mol.
• The molar mass of oxygen (O₂) is approximately 32.00 g/mol.
• The molar mass of nitric oxide (NO) is around 30.01 g/mol.

To perform calculations correctly:

• Convert the given mass of reactants to moles by dividing by their molar mass.
• Use these moles with the balanced chemical equation to determine the amounts of other substances produced or needed.

This step effectively links the equation's theoretical relationships to real-world quantities, facilitating accurate chemical reaction predictions and analyses.