Question

Balance this chemical equation by inspection: \(\mathrm{KClO}_{3} \rightarrow \mathrm{KCl}+\mathrm{O}_{2}\)

Short answer

To balance the given chemical equation \(\mathrm{KClO}_{3} \rightarrow \mathrm{KCl}+\mathrm{O}_{2}\), first balance the oxygens by adding a coefficient of 3/2 to \(\mathrm{O}_{2}\), resulting in \(\mathrm{KClO}_{3} \rightarrow \mathrm{KCl}+\frac{3}{2}\mathrm{O}_{2}\). Then, convert fractional coefficients to whole numbers by multiplying all coefficients by the least common denominator, which is 2 in this case. The balanced equation becomes \(2\mathrm{KClO}_{3} \rightarrow 2\mathrm{KCl} + 3\mathrm{O}_{2}\).

Step-by-step solution

Identify atoms on both sides

On the left side of the equation, we have one molecule of potassium chlorate (\(\mathrm{KClO}_{3}\)), containing: - 1 atom of potassium (K) - 1 atom of chlorine (Cl) - 3 atoms of oxygen (O) On the right side of the equation, we have one molecule of potassium chloride (\(\mathrm{KCl}\)) and one molecule of oxygen gas (\(\mathrm{O}_{2}\)), containing: - 1 atom of potassium (K) - 1 atom of chlorine (Cl) - 2 atoms of oxygen (O) Step 2: Balance oxygens by adding a coefficient to the \(\mathrm{O}_{2}\) term

Balance oxygens

We currently have 3 oxygen atoms on the left side of the equation and 2 on the right side. To balance the oxygens, we need to add a coefficient to the oxygen gas term so that there are equal oxygen atoms on both sides. In this case, we will add coefficient 3/2 to \(\mathrm{O}_{2}\), now we have: \[\mathrm{KClO}_{3} \rightarrow \mathrm{KCl}+\frac{3}{2}\mathrm{O}_{2}\] Step 3: Convert fractional coefficients to whole numbers.

Convert fractional coefficients

It's preferred to have whole number coefficients in a chemical equation. To achieve this, we can multiply the equation by the least common denominator (LCD) of all the coefficients. In this case, the LCD of the coefficients is 2. Multiplying all coefficients by the LCD, we get: \[ 2\times\mathrm{KClO}_{3} \rightarrow 2\times\mathrm{KCl} + 2\times\frac{3}{2}\mathrm{O}_{2} \] Which simplifies to: \[ 2\mathrm{KClO}_{3} \rightarrow 2\mathrm{KCl} + 3\mathrm{O}_{2} \] Now we have a balanced equation with whole number coefficients.

Key concepts

chemical reaction

A chemical reaction involves the transformation of one or more substances into different substances. This change occurs through the breaking and forming of chemical bonds. During a chemical reaction, the chemical and physical properties of the substances are altered, resulting in new products.
For instance, in our exercise involving potassium chlorate (\(\mathrm{KClO}_3\)), we observe a decomposition reaction. Decomposition reactions are a type of chemical reaction where one compound breaks down into two or more simpler substances. In this case, potassium chlorate decomposes into potassium chloride (\(\mathrm{KCl}\)) and oxygen gas (\(\mathrm{O}_2\)).
Understanding chemical reactions is pivotal as they describe how substances interact and change, forming the foundation for studying and applying chemistry across various disciplines.

• Reactants: Substances that start a reaction.
• Products: New substances formed as a result of the reaction.
• Decomposition: A type of chemical reaction where one compound splits into two or more substances.

stoichiometry

Stoichiometry is a branch of chemistry that deals with quantitatively analyzing the relationships between reactants and products in a chemical reaction. It helps in understanding the proportions of atoms involved and ensures that the law of conservation of mass holds true. This means that in any chemical reaction, matter is neither created nor destroyed.
The balancing of chemical equations is a critical application of stoichiometry. By ensuring that the number of atoms for each element is equal on both sides of the equation, we achieve a balanced equation. This involves determining the correct coefficients that will ensure this balance. These coefficients tell us the relative amounts of reactants and products and are crucial for calculating quantities in chemical reactions.
In the exercise, we initially balanced the oxygen atoms by adjusting the coefficient in front of \(\mathrm{O}_2\). Eventually, to avoid fractional coefficients, we simplified by multiplying all terms by 2. This resulted in a neat, whole number representation of the balanced equation.

• Balance: Equal number of atoms for each element in the reactants and products.
• Coefficients: Numbers placed in front of compounds to indicate their ratios in a reaction.
• Law of Conservation of Mass: Matter is preserved in any chemical transformation.

potassium chlorate decomposition

Potassium chlorate decomposition is a classic example of a decomposition reaction used frequently in chemistry education due to its straightforward nature and illustrative chemistry. When potassium chlorate (\(\mathrm{KClO}_3\)) is heated, it breaks down into potassium chloride (\(\mathrm{KCl}\)) and oxygen gas (\(\mathrm{O}_2\)). This reaction is significant for demonstrating how compounds can decompose into more stable elements or compounds.
Understanding this decomposition is crucial for learning how chemical bonds break and form, as well as the energy changes associated with reactions. In industrial processes, potassium chlorate decomposition can be used to produce oxygen, which is necessary for various applications.
The balanced chemical equation for the decomposition of potassium chlorate becomes:\[ 2\mathrm{KClO}_3 \rightarrow 2\mathrm{KCl} + 3\mathrm{O}_2 \]
This balanced equation reveals that for every 2 moles of potassium chlorate, 3 moles of oxygen gas are generated, exemplifying the practical applications of a balanced equation to predict product quantities in reactions.

• Heat: Often needed to drive decomposition reactions.
• Stable Products: Such as KCl and \(\mathrm{O}_2\), result from the reaction.
• Applications: Creates oxygen for industrial uses through controlled decomposition.