Question

A mixture containing ${\mathrm{KCIO}}_3,{\mathrm{K}}_2{\mathrm{CO}}_3,{\mathrm{KHCO}}_3$, and $\mathrm{KCl}$ was heated, producing ${\mathrm{CO}}_2,{\mathrm{O}}_2$, and ${\mathrm{H}}_2\mathrm{O}$ gases according to the following equations: $$\begin{array}{rl}2{\mathrm{KClO}}_3(\mathrm{s})& ⟶2\mathrm{KCl}(s)+3{\mathrm{O}}_2(g)\\ 2{\mathrm{KHCO}}_3(\mathrm{s})& ⟶{\mathrm{K}}_2\mathrm{O}(s)+{\mathrm{H}}_2\mathrm{O}(g)+2{\mathrm{CO}}_2(g)\\ {\mathrm{K}}_2{\mathrm{CO}}_3(s)& {\mathrm{K}}_2\mathrm{O}(s)+{\mathrm{CO}}_2(g)\end{array}$$ The KCl does not react under the conditions of the reaction. If $100.0\mathrm{g}$ of the mixture produces $1.80\mathrm{g}$ of ${\mathrm{H}}_2\mathrm{O}$, $13.20\mathrm{g}$ of ${\mathrm{CO}}_2$, and $4.00\mathrm{g}$ of ${\mathrm{O}}_2$, what was the composition of the original mixture? (Assume complete decomposition of the mixture.)

Short answer

The composition of the original 100.0 g mixture was as follows: - 12.98 g of KCIO₃ - 5.01 g of KHCO₃ - 27.64 g of K₂CO₃ - 54.37 g of KCl

Step-by-step solution

Identify the molar masses of the produced gases

We need to find the molar masses of H₂O, CO₂, and O₂ in order to get their moles. The molar masses are given as the sum of the individual atomic masses: - H₂O: 2(1.008) + 16.00 = 18.016 g/mol - CO₂: 12.01 + 2(16.00) = 44.01 g/mol - O₂: 2(16.00) = 32.00 g/mol

Calculate the moles of the produced gases

Now that we have the molar masses of each gas, we will use the given masses produced during the reaction to find the moles of H₂O, CO₂, and O₂. - Moles of H₂O: 1.80 g / 18.016 g/mol = 0.1 mol - Moles of CO₂: 13.20 g / 44.01 g/mol = 0.3 mol - Moles of O₂: 4.00 g / 32.00 g/mol = 0.125 mol

Determine the amounts of the original compounds in moles

Using the balanced equations and stoichiometry, we will find the moles of the original compounds that produced those gases. From the balanced equations: KCIO₃ ⟶ O₂: 2 moles of KCIO₃ ⟶ 3 moles of O₂ KHCO₃ ⟶ H₂O: 2 moles of KHCO₃ ⟶ 1 mole of H₂O K₂CO₃ and KHCO₃ ⟶ CO₂: 1 mole of K₂CO₃ + 2 moles of KHCO₃ ⟶ 3 moles of CO₂ Now we have a system of equations with the moles of each compound (x, y, and z): $$2x=3(0.125)$$ $$2y=0.1$$ $$z+2y=0.3$$ Solving these equations, we get: x = 0.09375 mol (of KCIO₃) y = 0.05 mol (of KHCO₃) z = 0.2 mol (of K₂CO₃)

Calculate the masses of the original compounds

Now that we have the moles of each compound, we will multiply them by their respective molar masses to find the masses of each. - Mass of KCIO₃: 0.09375 mol * 138.55 g/mol = 12.98 g - Mass of KHCO₃: 0.05 mol * 100.12 g/mol = 5.01 g - Mass of K₂CO₃: 0.2 mol * 138.21 g/mol = 27.64 g Recalling that the total mass of the original mixture should be 100 g, we can find the mass of KCl in the original mixture: - Mass of KCl = 100 g - (12.98 g + 5.01 g + 27.64 g) = 54.37 g

Find the composition of the original mixture

Now that we have the weights of each compound, we can determine the composition of the original mixture containing KCIO₃, KHCO₃, K₃CO₃, and KCl. The original mixture was composed of: - 12.98 g of KCIO₃ - 5.01 g of KHCO₃ - 27.64 g of K₂CO₃ - 54.37 g of KCl

Key concepts

Stoichiometry

Stoichiometry is like a recipe in chemistry. It helps us understand the relationships between different reactants and products in a chemical reaction.
Using stoichiometry, we can predict how much of a product is formed from given amounts of reactants. This occurs by using balanced chemical equations.
In our example, we use stoichiometry to determine the moles of gases produced from the decomposition of the mixture.

• For each reaction, the coefficients in the balanced equation indicate the proportion of reactants to products.


• This provides a mole ratio, which is a key part of stoichiometric calculations.

Understanding stoichiometry allows us to work backwards, using the amount of gas produced to find the original composition of the mixture.

Molar Mass

Molar mass is the mass of one mole of any given substance. It is very important in stoichiometry.When dealing with chemical reactions, being able to convert between mass and moles is essential. This requires knowing the molar mass of each component.
- The molar mass of a compound is calculated by summing the atomic masses of all the atoms in a molecule.For example:

• Water (${\mathrm{H}}_2\mathrm{O}$): 2 hydrogen atoms and 1 oxygen atom. $2\times 1.008+16.00=18.016\mathrm{g}/\mathrm{mol}$.


• Carbon dioxide (${\mathrm{CO}}_2$): 1 carbon atom and 2 oxygen atoms. $12.01+2\times 16.00=44.01\mathrm{g}/\mathrm{mol}$.


• Oxygen (${\mathrm{O}}_2$): 2 oxygen atoms. $2\times 16.00=32.00\mathrm{g}/\mathrm{mol}$.

Knowing molar mass is crucial for calculating the moles of a substance from its mass, which is the basis for further stoichiometric calculations.

Chemical Reactions

Chemical reactions involve transforming substances into different substances by breaking and forming bonds between atoms.
In this exercise, we're looking at decomposition reactions, where compounds break down into simpler products, usually gases in this case.
The reactions given in the exercise are all examples of decomposition:

• $2{\mathrm{KClO}}_3\to 2\mathrm{KCl}+3{\mathrm{O}}_2$


• $2{\mathrm{KHCO}}_3\to {\mathrm{K}}_2\mathrm{O}+{\mathrm{H}}_2\mathrm{O}+2{\mathrm{CO}}_2$


• ${\mathrm{K}}_2{\mathrm{CO}}_3\to {\mathrm{K}}_2\mathrm{O}+{\mathrm{CO}}_2$

These equations provide a roadmap, detailing how much of each product forms from a given amount of reactant.
Understanding these reactions involves identifying which compounds break down and the nature of the products produced.

Gas Production

In many chemical reactions, especially those involving decomposition, gases are a common product. The measurement of these gases can tell us a lot.
Gas production, in our example, involves the release of oxygen, carbon dioxide, and water vapor. Knowing how much gas is produced helps us trace the path of the reaction.
The produced gases often are:

• Oxygen (${\mathrm{O}}_2$): Key for reactions, but also accounts for volumetric and mass changes in experiments.


• Carbon Dioxide (${\mathrm{CO}}_2$): Commonly seen in combustion and decomposition.


• Water Vapor (${\mathrm{H}}_2\mathrm{O}$): Often a product in many chemical reactions.

In analytical chemistry, the amount of gas produced can be used to backtrack and find the starting amounts of other substances.
By calculating the moles of gas produced, you can work out the stoichiometry of the original components in the mixture.