If \(250 \mathrm{~mL}\) of \(\mathrm{O}_{2}\) measured at STP, are obtained by the decomposition of the \(\mathrm{KClO}_{3}\) in a \(1.20\)-g mixture of \(\mathrm{KCl}\) and \(\mathrm{KClO}_{3}\), $$ 2 \mathrm{KClO}_{3}(s) \longrightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g) $$ what is the percent by mass of \(\mathrm{KClO}_{3}\) in the mixture?

Understanding stoichiometry is essential for solving problems involving the quantitative aspects of chemical reactions. It is the study of the relative quantities of reactants and products in chemical reactions and is foundational for many topics in chemistry, including determining percent composition by mass.

At its core, stoichiometry requires a balanced chemical equation. Every balanced equation provides a proportion—known as the stoichiometric ratio—between the molecules or moles of reactants and products. For instance, the given chemical reaction decomposition for the exercise:
\[ 2 \mathrm{KClO}_{3}(s) \longrightarrow 2 \mathrm{KCl}(s) + 3 \mathrm{O}_{2}(g) \]
shows the stoichiometric relationship that two moles of potassium chlorate \(\mathrm{KClO}_{3}\) yield three moles of oxygen \(\mathrm{O}_{2}\) gas upon decomposition. This ratio is critical when you want to find out how much reactant is needed to produce a certain amount of product, or vice versa, which is precisely what the problem asks for.

The molar volume of a gas at standard temperature and pressure (STP), which is 0°C and 1 atmospheric pressure, is a physical constant that proofs invaluable for various calculations in chemistry. Specifically, the molar volume at STP is 22.4 liters per mole (L/mol) or 22400 milliliters per mole (mL/mol), applicable to any ideal gas.

This constant allows us to convert between the volume of a gas and the number of moles of the gas when the gas is at STP. The exercise provided utilizes this concept: Given 250 mL of \(\mathrm{O}_{2}\) at STP, one can find the number of moles of \(\mathrm{O}_{2}\) using the equation:
\[ \text{Moles of } \mathrm{O}_{2} = \frac{Volume \ of \ \mathrm{O}_{2} \(mL\)}{Molar \ volume \(mL/mol\)} \].
Applying this equation allows us to then relate the volume of gas produced to the amount of reactant used, as we use stoichiometry to connect the two.

Decomposition reactions are a type of chemical reaction where one compound breaks down into two or more simpler substances. In our exercise, the decomposition of potassium chlorate \(\mathrm{KClO}_{3}\) into potassium chloride \(\mathrm{KCl}\) and oxygen gas \(\mathrm{O}_{2}\) is an example of such a reaction.

Understanding these reactions requires a grasp of the stoichiometry involved as well as the ability to apply concepts such as molar volume at STP to gaseous products. The decomposition equation provided is crucial for identifying the number of moles of each substance that participate in the reaction. Once the number of moles of oxygen gas is calculated from the given volume, we can use the balanced equation to determine the moles, and thus the mass, of the potassium chlorate that underwent decomposition. This ultimately leads us to calculate percent composition by mass—a key step to assess the purity or the composition of a given mixture in chemistry.

Tip:

• Always verify that the chemical equation is balanced before using it to relate the amounts of reactants and products in a decomposition reaction.