Question

Potassium nitrite can be produced by heating a mixture of potassium nitrate and carbon. Write a balanced equation for this reaction. Calculate the theoretical yield of \(\mathrm{KNO}_{2}\) produced by heating \(57.0 \mathrm{~g}\) of \(\mathrm{KNO}_{3}\) with an excess of carbon.

Short answer

48.0 g of KNO₂ is produced.

Step-by-step solution

Write the Unbalanced Chemical Equation

Start by writing the unbalanced chemical equation for the decomposition of potassium nitrate (\(\text{KNO}_3\)) in the presence of carbon. The reaction is: \[\text{KNO}_3 + \text{C} \rightarrow \text{KNO}_2 + \text{CO}_2\]

Balance the Chemical Equation

Balance the chemical equation by adjusting the coefficients to have the same number of each type of atom on both sides of the equation:\[2\text{KNO}_3 + \text{C} \rightarrow 2\text{KNO}_2 + \text{CO}_2\]Each side now has 2 K, 2 N, 6 O, and 1 C atom.

Calculate Molar Masses

Calculate the molar mass of each compound involved:- \(\text{KNO}_3\) has a molar mass of \(39.1 + 14.0 + (16.0 \times 3) = 101.1\,\text{g/mol}\).- \(\text{KNO}_2\) has a molar mass of \(39.1 + 14.0 + (16.0 \times 2) = 85.1\,\text{g/mol}\).

Determine Moles of Reactant

Convert \(57.0\,\text{g}\) of \(\text{KNO}_3\) to moles:\[\text{moles of KNO}_3 = \frac{57.0\,\text{g}}{101.1\,\text{g/mol}} \approx 0.564\,\text{mol}\]

Use Stoichiometry to Find Moles of Product

Use the stoichiometry from the balanced equation to find the moles of \(\text{KNO}_2\):For every 2 moles of \(\text{KNO}_3\), 2 moles of \(\text{KNO}_2\) are formed. Therefore, the moles of \(\text{KNO}_2\) from \(0.564\) moles of \(\text{KNO}_3\) is also \(0.564\) moles.

Calculate Theoretical Yield of KNO₂

Convert moles of \(\text{KNO}_2\) to grams to find the theoretical yield:\[\text{grams of KNO}_2 = 0.564\,\text{mol} \times 85.1\,\text{g/mol} = 48.0\,\text{g}\]

Conclusion

The balanced equation is \(2\text{KNO}_3 + \text{C} \rightarrow 2\text{KNO}_2 + \text{CO}_2\) and the theoretical yield of \(\text{KNO}_2\) is \(48.0\,\text{g}\).

Key concepts

Chemical Equation

A chemical equation is a symbolic representation of a chemical reaction. It shows the reactants and products, as well as their quantities. In the exercise, the initial unbalanced chemical equation was:

• Reactants: Potassium nitrate (\(\text{KNO}_3\)) and Carbon (\(\text{C}\))
• Products: Potassium nitrite (\(\text{KNO}_2\)) and Carbon dioxide (\(\text{CO}_2\))

Writing a chemical equation correctly is the first step in solving stoichiometry problems. Each molecule and its respective state (solid, liquid, gas, or aqueous) must be specified, although states are not always detailed in introductory problems. Understanding the chemical equation's structure aids in visualizing and balancing each component of the reaction.

Theoretical Yield

Theoretical yield represents the maximum amount of product that can be formed from a given amount of reactants, based on the stoichiometric relationships in a balanced chemical equation. It assumes that the reaction goes to completion and that no products are lost during the process. In the exercise, the theoretical yield of \(\text{KNO}_2\) was calculated using stoichiometric conversions between mass, molar mass, and moles.

To determine the theoretical yield, follow these steps:

• Convert the mass of reactants to moles using their molar masses.
• Use the stoichiometric relationships in the balanced equation to calculate the moles of the desired product.
• Finally, convert the moles of product back to grams using its molar mass.

By doing this, the maximum possible amount of \(\text{KNO}_2\) was found to be 48.0 grams when using 57.0 grams of \(\text{KNO}_3\).

Molar Mass

Molar mass is the mass of one mole of a substance, expressed in grams per mole. It is crucial in stoichiometric calculations as it allows you to convert between mass and moles. In the exercise, the molar masses needed were:

• For \(\text{KNO}_3\): Potassium (K) 39.1 g/mol, Nitrogen (N) 14.0 g/mol, Oxygen (O) 16.0 g/mol. Addition yields a molar mass of 101.1 g/mol.
• For \(\text{KNO}_2\): Potassium (K) 39.1 g/mol, Nitrogen (N) 14.0 g/mol, and Oxygen (O) 16.0 g/mol. Together, this gives a molar mass of 85.1 g/mol.

Calculating molar masses requires adding the atomic masses of each element in the molecule according to its chemical formula. These calculations are vital for converting between the mass of reactants or products and the amount of substance, commonly measured in moles.

Balancing Chemical Equations

Balancing chemical equations ensures that the same number of each type of atom appears on both sides of the equation. This step is crucial because it respects the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction.

In the exercise, the unbalanced equation was \(\text{KNO}_3 + \text{C} \rightarrow \text{KNO}_2 + \text{CO}_2\). By balancing it, every element had to have an equal count on each side:

• 2 Potassium atoms on both sides
• 2 Nitrogen atoms on both sides
• 6 Oxygen atoms on both sides
• 1 Carbon atom on both sides

The balanced equation, \(2\text{KNO}_3 + \text{C} \rightarrow 2\text{KNO}_2 + \text{CO}_2\), provides the correct stoichiometric coefficients. This balancing is necessary to correctly calculate the amount of products or reactants in any stoichiometric problem.