Question

Oxygen gas can be produced in the laboratory by decomposition of potassium chlorate \(\left(\mathrm{KClO}_{3}\right)\) $$ \mathrm{KClO}_{3}(s) \longrightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g) $$ How many \(\mathrm{kg}\) of \(\mathrm{O}_{2}\) can be produced from \(1.0 \mathrm{~kg}\) of \(\mathrm{KClO}_{3} ?\)

Short answer

Approximately 0.39 kg of O2 can be produced from 1.0 kg of KClO3.

Step-by-step solution

Calculate the molar mass of KClO3

Using the periodic table, find the atomic masses of potassium (K), chlorine (Cl), and oxygen (O) and add them together to get the molar mass of KClO3. The molar mass of KClO3 is the sum of the molar masses of its constituent atoms: molar mass of K (39 g/mol) + molar mass of Cl (35.45 g/mol) + 3 * molar mass of O (3 * 16 g/mol) = 122.55 g/mol.

Calculate the molar mass of O2

Determine the molar mass of O2 which is the mass of two oxygen atoms. Since the atomic mass of oxygen is 16 g/mol, the molar mass of O2 is 2 * 16 g/mol = 32 g/mol.

Use stoichiometry to find the moles of O2 produced from KClO3

According to the balanced equation, 2 moles of KClO3 produce 3 moles of O2. First, convert the mass of KClO3 to moles by dividing by its molar mass (found in Step 1). Then use the mole ratio from the balanced equation to find the moles of O2 produced from KClO3.

Convert moles of O2 into grams

Multiply the moles of O2 found in Step 3 by the molar mass of O2 to obtain the mass of O2 in grams.

Convert grams of O2 into kilograms

Convert the mass of O2 in grams to kilograms by dividing by 1000 since there are 1000 grams in a kilogram.

Key concepts

Stoichiometry

Stoichiometry is the section of chemistry that involves using balanced chemical equations to calculate the relative quantities of reactants and products involved in chemical reactions. When potassium chlorate \(\mathrm{KClO}_{3}\) decomposes, it produces potassium chloride \(\mathrm{KCl}\) and oxygen gas \(\mathrm{O}_{2}\). The balanced equation \(\mathrm{KClO}_{3}(s) \longrightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)\) tells us that for every one mole of \(\mathrm{KClO}_{3}\) used, three moles of \(\mathrm{O}_{2}\) are produced.

To fully understand stoichiometry, students need to grasp the concept of the mole, which is a unit for counting atoms and molecules. One mole is equivalent to Avogadro's number \(6.022 \times 10^{23}\) entities. Stoichiometry allows us to predict the amounts of products formed in a chemical reaction based on the amounts of reactants used.

Molar Mass Calculation

Molar mass is a fundamental concept in stoichiometry as it links the mass of a substance to the amount in moles. The molar mass of a compound is the sum of the masses of all the atoms in one mole of the compound. For example, the molar mass of potassium chlorate \(\mathrm{KClO}_{3}\) is calculated by adding the atomic masses of potassium \(K\), chlorine \(Cl\), and oxygen \(O\) together:

• Molar mass of \(K\): 39 g/mol
• Molar mass of \(Cl\): 35.45 g/mol
• Molar mass of \(O\) (times 3, since there are three oxygen atoms): 3 * 16 g/mol

In the provided exercise, knowing the molar mass of \(\mathrm{KClO}_{3}\) and \(\mathrm{O}_{2}\) is key to converting between mass and moles, allowing students to traverse the bridge from the reactant's mass to the product's mass.

Chemical Reaction

A chemical reaction is a process in which one or more substances, the reactants, are converted into one or more different substances, the products. Understanding the nature of chemical reactions includes recognizing that mass is conserved throughout the process. In the decomposition of potassium chlorate, the reactant \(\mathrm{KClO}_{3}\) undergoes a chemical change to form new substances, potassium chloride \(\mathrm{KCl}\) and oxygen gas \(\mathrm{O}_{2}\).

Chemical reactions are represented by balanced chemical equations, which demonstrate the conservation of atoms. Each atom on the reactant side must be accounted for on the product side, as shown in the balanced equation for this reaction. It's essential to comprehend that the coefficients in a balanced equation indicate the molar ratios of reactants and products involved. This fundamental principle is utilized in stoichiometric calculations to precisely determine the quantities of each substance required or produced.

Theoretical Yield

The theoretical yield of a chemical reaction is the maximum amount of product that could be formed from the given quantities of reactants, assuming perfect conversion and no losses. It is calculated using the balanced chemical equation and the stoichiometry of the reaction. In the decomposition of \(\mathrm{KClO}_{3}\), you first determine the number of moles of \(\mathrm{KClO}_{3}\) you're starting with by dividing its mass by its molar mass. Next, using the stoichiometry of the reaction, you find out how many moles of \(\mathrm{O}_{2}\) would be produced from those moles of \(\mathrm{KClO}_{3}\). The theoretical yield in grams is found by multiplying this number by the molarfab mass of \(\mathrm{O}_{2}\). Finally, to find the theoretical yield in kilograms, which was the question in the exercise, you convert the grams into kilograms.

The concept of theoretical yield is crucial for planning reactions and understanding the efficiency of a process. In practice, the actual yield is often less than the theoretical yield due to side reactions, incomplete reactions, or other practical factors in the laboratory.