Select the correct alternative from the given choices. \(0.1225 \mathrm{~g}\) of potassium chlorate decomposes completely. Write the possible chemical equation and the amount of oxygen liberated. (a) \(4 \mathrm{KClO}_{3} \rightarrow 4 \mathrm{KCl}+6 \mathrm{O}_{2}, 0.048 \mathrm{~g}\) (b) \(4 \mathrm{KClO}_{3} \rightarrow 4 \mathrm{KCl}+5 \mathrm{O}_{2}, 0.48 \mathrm{~g}\) (c) \(2 \mathrm{KClO}_{3} \rightarrow 2 \mathrm{KCl}+3 \mathrm{O}_{2}, 0.048 \mathrm{~g}\) (d) \(2 \mathrm{KClO}_{3} \rightarrow 2 \mathrm{KCl}+3 \mathrm{O}_{2}, 0.48 \mathrm{~g}\)

A balanced chemical equation is fundamentally important in the realm of chemistry. It represents a reaction where the number of atoms for each element is equal on both sides, adhering to the law of conservation of mass. This means that atoms aren't created or destroyed in a chemical reaction; they are simply rearranged.

To achieve a balanced equation, coefficients are placed in front of the chemical formulas to ensure the same number of atoms of each element appear on both sides of the equation. In our potassium chlorate decomposition example, the balanced version is indicated as option (c) with the equation: \(2KClO_3 \rightarrow 2KCl + 3O_2\), where we see two moles of potassium chlorate yielding two moles of potassium chloride and three moles of oxygen gas.

The molar mass is the weight of one mole of a substance, usually expressed in grams per mole (g/mol). It is a critical factor in stoichiometry as it connects the number of particles, like atoms or molecules, to the mass of a sample. To calculate the molar mass, you sum up the atomic masses of all the atoms in the molecule. For instance, for potassium chlorate (KClO3), you add the atomic masses of one potassium (K), one chlorine (Cl), and three oxygen (O) atoms, resulting in \(39.1 + 35.5 + 3 \times 16 = 122.55g/mol\).

It is these calculations that enable us to convert between the mass of a substance and the number of moles, paving the way for further stoichiometric conversions, such as finding out how much of one substance can react with a given amount of another substance.

Stoichiometric calculations are the quantitative aspect of chemical equations, allowing chemists to predict the amounts of reactants needed and products formed in a chemical reaction. Stoichiometry uses the balanced chemical equation as a basis to perform these calculations.

In the problem at hand, stoichiometry is applied to decipher the mass of oxygen gas released when a specific mass of potassium chlorate decomposes. We start by converting the mass of potassium chlorate to moles using its molar mass: \(0.1225g \/ 122.55g/mol = 0.001mol\). We then apply the mole ratio from the balanced equation to ascertain the moles of oxygen produced. Multiplying this by the molar mass of oxygen (32 g/mol) gives us the mass of oxygen gas produced from the given mass of potassium chlorate. This systematic approach is the cornerstone of stoichiometric calculations and emphasizes the importance of understanding molar mass and balanced equations.