Question

A chemist has a mixture of \(\mathrm{KClO}_{3}, \mathrm{KHCO}_{3}, \mathrm{~K}_{2} \mathrm{CO}_{3}\), and \(\mathrm{KCl}\). She heats \(1,000 \mathrm{~g}\) of this mixture and notices that the following gases evolve: 18 g of water \(\left(\mathrm{H}_{2} \mathrm{O}\right), 132 \mathrm{~g}\) of \(\mathrm{CO}_{2}\), and \(40 \mathrm{~g}\) of \(\mathrm{O}_{2}\) according to the following reactions: $$ \begin{array}{ll} 2 \mathrm{KC} 1 \mathrm{O}_{3} & \rightarrow 2 \mathrm{KCl}+3 \mathrm{O}_{2} \\\ 2 \mathrm{KHCO}_{3} & \rightarrow \mathrm{K}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O}+2 \mathrm{CO}_{2} \\ \mathrm{~K}_{2} \mathrm{CO}_{3} & \rightarrow \mathrm{K}_{2} \mathrm{O}+\mathrm{CO}_{2} \end{array} $$ The \(\mathrm{KCl}\) is inert under these conditions. Assuming complete decomposition, determine the composition of the original mixture.

Short answer

The composition of the original mixture is: \(\mathrm{KClO}_3\): 102.09g \(\mathrm{KHCO}_3\): 200.24g \(\mathrm{K_2CO}_3\): 138.21g \(\mathrm{KCl}\): 559.46g

Step-by-step solution

List the compounds and their masses

The problem tells us that we have a mixture of 1000g and that 18g of water, 40g of oxygen and 132g of carbon dioxide are produced. These quantities will help us find the initial composition of the mixture.

Use stoichiometry to determine the composition of the original mixture

Since we know the masses of produced gases, we can use the stoichiometry of each reaction to determine the amount of each compound that has reacted. 1. For \(\mathrm{O}_2\): From the first reaction, 2 moles of \(\mathrm{KClO}_3\) produce 3 moles of \(\mathrm{O}_2\). The molar mass of \(\mathrm{KClO}_3\) is \(39.1 + 35.45 + 3 \times 16 = 122.55 \mathrm{~g/mol}\), and \(\mathrm{O}_2\) is \(2 \times 16 = 32 \mathrm{~g/mol}\). Now, \(40 \mathrm{~g}\) of \(\mathrm{O}_2\) is produced, let's find the moles: \[\frac{40 \mathrm{~g}}{32 \mathrm{~g/mol}} = 1.25 \mathrm{~mol}\] Using stoichiometry, the moles of \(\mathrm{KClO}_3\) reacted is: \[\frac{2}{3} \times 1.25 \mathrm{~mol} = 0.833 \mathrm{~mol}\] Hence, the mass of \(\mathrm{KClO}_3\) reacted is: \(0.833 \mathrm{~mol} \times 122.55 \mathrm{~g/mol} = 102.09 \mathrm{~g}\). 2. For \(\mathrm{CO}_2\): From the second and third reactions, we know that 2 moles of \(\mathrm{KHCO}_3\) produce 2 moles of \(\mathrm{CO}_2\), and 1 mole of \(\mathrm{K_2CO}_3\) produce 1 mole of \(\mathrm{CO}_2\). The molar mass of \(\mathrm{KHCO}_3\) is \(39.1 + 1.01 + 12.01 + 3 \times 16 = 100.12 \mathrm{~g/mol}\) and \(\mathrm{K_2CO}_3\) is \(2 \times 39.1 + 12.01 + 3 \times 16 = 138.21 \mathrm{~g/mol}\). With \(132 \mathrm{~g}\) of \(\mathrm{CO}_2\), moles of \(\mathrm{CO}_2\): \[\frac{132 \mathrm{~g}}{44.01 \mathrm{~g/mol}} = 3 \mathrm{~mol}\] Let's assume that x moles are generated by reaction 2, then \(3-x\) moles are generated by reaction 3. So, we have: Mass of \(\mathrm{KHCO}_3\) = \(2x \times 100.12 \mathrm{~g/mol}\) Mass of \(\mathrm{K_2CO}_3\) = \((3-x) \times 138.21 \mathrm{~g/mol}\) 3. For \(\mathrm{H_2O}\): From reaction 2, 2 moles of \(\mathrm{KHCO}_3\) produce 1 mole of \(\mathrm{H_2O}\). With \(18 \mathrm{~g}\) of \(\mathrm{H_2O}\), moles of \(\mathrm{H_2O}\): \[\frac{18 \mathrm{~g}}{18.02 \mathrm{~g/mol}} = 1 \mathrm{~mol}\] Using stoichiometry, the moles of \(\mathrm{KHCO}_3\) reacted in the second reaction are: \(2 \times 1 \mathrm{~mol} = 2 \mathrm{~mol}\). Hence, mass of \(\mathrm{KHCO}_3 = 2 \times 100.12 \mathrm{~g/mol} = 200.24 \mathrm{~g}\)\). Since we already know that the moles of \(\mathrm{KHCO}_3\) reacted are 2, the moles of \(\mathrm{K_2CO}_3\) reacted in reaction 3 are \(3-2=1\) mole. Thus, mass of \(\mathrm{K_2CO}_3 = 1 \times 138.21 \mathrm{~g/mol} = 138.21 \mathrm{~g}\)\).

Calculate the mass of unreacted KCl

The mass of the original mixture is 1000g, so now we can calculate the mass of unreacted \(\mathrm{KCl}\): Mass of \(\mathrm{KCl} = 1000 \mathrm{~g} - 102.09 \mathrm{~g} - 200.24 \mathrm{~g} - 138.21 \mathrm{~g} = 559.46 \mathrm{~g}\) So, the composition of the original mixture is: \(\mathrm{KClO}_3\): 102.09g \(\mathrm{KHCO}_3\): 200.24g \(\mathrm{K_2CO}_3\): 138.21g \(\mathrm{KCl}\): 559.46g

Key concepts

Chemical Reactions

Chemical reactions are processes where substances transform into different ones. In the context of this problem, several reactants decompose to release specific gases. For example, \(\mathrm{KClO}_3\) undergoes decomposition to form \(\mathrm{KCl}\) and oxygen gas \(\mathrm{O}_2\). The second reaction involves \(\mathrm{KHCO}_3\) decomposing into water \(\mathrm{H}_2\mathrm{O}\), carbon dioxide \(\mathrm{CO}_2\), and potassium oxide \(\mathrm{K_2O}\). Lastly, \(\mathrm{K_2CO}_3\) decomposes into carbon dioxide \(\mathrm{CO}_2\) and potassium oxide \(\mathrm{K_2O}\). Important points to remember about chemical reactions include:

• Reactants are substances that undergo change.
• Products are new substances formed as a result of the reaction.
• Balance equations to ensure that the mass and the number of atoms are conserved.

These reactions illustrate the decomposition type, where complex molecules break down into simpler substances. To solve problems involving reactions, balancing the equation helps keep track of the atoms and molecules involved.

Molar Mass

Molar mass is the mass of one mole of a substance. It is expressed in grams per mole (g/mol) and is used to convert between grams and moles in stoichiometric calculations. It can be found by adding the atomic masses of all atoms in a molecule, as shown in the periodic table.In this exercise, calculating the molar masses was essential:

• \(\mathrm{KClO}_3\): The molar mass is calculated as \(39.1 + 35.45 + 3 \times 16 = 122.55 \, \mathrm{g/mol}\).
• \(\mathrm{KHCO}_3\): Its molar mass is \(39.1 + 1.01 + 12.01 + 3 \times 16 = 100.12 \, \mathrm{g/mol}\).
• \(\mathrm{K_2CO}_3\): The molar mass is \(2 \times 39.1 + 12.01 + 3 \times 16 = 138.21 \, \mathrm{g/mol}\).

Knowing these values allows us to convert the mass of each compound to moles, facilitating calculations necessary to find the composition of a mixture. Molar masses are powerful tools used in stoichiometry to relate mass to number of atoms or molecules.

Mass Composition

Mass composition refers to the proportion of each substance in a mixture based on its mass. In stoichiometry, this involves determining how much of each reactant is present initially and how it changes after a reaction. Here, the total mass of gases evolved helps us deduce the amount and type of substances initially present in the mixture.To determine mass composition:

• Calculate the total mass of all products to verify the decomposition extent.
• Subtract the mass of gaseous products from the initial mixture to find unreacted substances, such as \(\mathrm{KCl}\).
• Assign masses to each reactant using stoichiometric ratios obtained from balanced reactions.

In the given problem, identifying

• 18 g of water (\(\mathrm{H}_2\mathrm{O}\))
• 40 g of oxygen (\(\mathrm{O}_2\))
• 132 g of carbon dioxide (\(\mathrm{CO}_2\))

provided clues on how much of \(\mathrm{KClO}_3\), \(\mathrm{KHCO}_3\), and \(\mathrm{K_2CO}_3\) decomposed. This approach ensures that all components are accounted for accurately.

Gas Evolution

Gas evolution is a key concept in reactions where gases are produced as products. Understanding which gases evolve during reactions helps determine the progression and completion of the reactions. In the context of this exercise, three different gases evolve, indicating decomposition reactions.The gases evolved in these reactions are derived from reactions involving decomposition of different potassium compounds:

• \(\mathrm{KClO}_3\) produces \(\mathrm{O}_2\) gas.
• \(\mathrm{KHCO}_3\) produces both \(\mathrm{H}_2\mathrm{O}\) and \(\mathrm{CO}_2\).
• \(\mathrm{K_2CO}_3\) contributes to \(\mathrm{CO}_2\) evolution.

The measured quantities of evolved gases provide analytical data to back-calculate and find out how much of each reactant was initially present. Gas evolution often indicates that decomposition reactions have taken place successfully, shedding light on the dynamics of chemical processes within a mixture.