Oxygen gas can be generated by the decomposition of potassium chlorate according to the following equation: $$ 2 \mathrm{KClO}_{3}(s) \longrightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g) $$ How many liters of oxygen at STP will be produced when \(0.525 \mathrm{~kg}\) of \(\mathrm{KCl}\) is also produced?

Understanding stoichiometry calculations is essential to solving many chemistry-related problems. Stoichiometry is the branch of chemistry that deals with the quantitative relationships or ratios between reactants and products in a chemical reaction.

In our exercise, we observed stoichiometry in action when we used the balanced chemical equation to relate the moles of potassium chlorate (KClO₃) decomposing to produce potassium chloride (KCl) and oxygen gas (O₂). Each mole of a substance contains Avogadro's number of representative particles, such as atoms, molecules, or formula units, enabling us to convert between moles and mass.

Stoichiometry is akin to a recipe—in this case, the equation tells us that 2 moles of KClO₃ yield 2 moles of KCl and 3 moles of O₂. Like a chef converting cups to tablespoons, a chemist must convert between moles of different substances using the mole ratios from the balanced equation.

The molar mass determination is a pivotal skill in chemistry, as it allows you to convert between the mass of a substance and moles, which is a fundamental stepping stone for many different types of calculations. It is the mass of one mole of that substance—the average mass of all isotopes of all atoms present in the molecule, accounting for their natural abundance, measured in grams per mole (g/mol).

In the exercise, we determined molar mass by summing the atomic masses of each element that composes the compound, which are provided on the periodic table. For KClO₃, the calculations were (39.10 g/mol for K) + (35.45 g/mol for Cl) + (3 x 16.00 g/mol for O), while for KCl, it was the sum of the atomic masses of potassium and chlorine. Once the molar mass was found, we could convert the given mass of KCl from the exercise into moles, enabling us to proceed with the stoichiometry calculations.

Applying the Ideal Gas Law at Standard Temperature and Pressure (STP) is a crucial component for calculating gaseous volumes in chemical reactions. STP is defined as a temperature of 0 degrees Celsius (273.15 Kelvin) and a pressure of 1 atmosphere. At these conditions, one mole of any ideal gas occupies 22.4 liters, a fact that greatly simplifies calculations involving gases.

In our solution, we used the stoichiometry to find the moles of O₂ produced and then used the property of gases at STP to find the volume. By multiplying the moles of oxygen by 22.4 L/mol, we were able to quickly calculate the volume of oxygen that would be produced from our reaction at STP. This step is an application of Avogadro's Law, which states that equal volumes of ideal gases, at the same temperature and pressure, have the same number of molecules, and is a cornerstone of the ideal gas law expressed as PV=nRT with 'n' being the moles of gas.