Question

Classify the type of stoichiometry problem: How many grams of potassium chlorate must decompose to give \(1.00 \mathrm{~g}\) of \(\mathrm{KCl}\) ?

Short answer

It's a mass-to-mass stoichiometry problem.

Step-by-step solution

Understand the Reaction

The decomposition of potassium chlorate (KClO_3) is given by the reaction:\[2 ext{KClO}_3 ightarrow 2 ext{KCl} + 3 ext{O}_2\]This reaction shows that 2 moles of KCl are produced from 2 moles of KClO_3.

Calculate Moles of KCl

First, we need to calculate the number of moles in 1.00 g of KCl. The molar mass of KCl is approximately 74.55 g/mol. Using the formula:\[ ext{moles of KCl} = \frac{1.00 ext{ g}}{74.55 ext{ g/mol}} \]this gives approximately 0.0134 moles of KCl.

Use Reaction Stoichiometry

According to the balanced chemical equation, the stoichiometry between KClO_3 and KCl is 1:1. Therefore, 0.0134 moles of KClO_3 are required to produce 0.0134 moles of KCl.

Calculate Mass of KClO3

Now, calculate the mass of KClO_3 needed using its molar mass (approx. 122.55 g/mol):\[ ext{mass of KClO}_3 = 0.0134 ext{ moles} \times 122.55 ext{ g/mol} = 1.643 ext{ g}\]

Classify Type of Stoichiometry Problem

This is a mass-to-mass stoichiometry problem, involving conversion from grams of product to grams of reactant using a balanced chemical equation.

Key concepts

Chemical Reactions

Chemical reactions involve the transformation of reactants into products through a process that alters their chemical composition. In the reaction where potassium chlorate (KClO_3) decomposes into potassium chloride (KCl) and oxygen (O_2), we witness this transformative process firsthand. The decomposition reaction can be represented as:\[ 2 \text{KClO}_3 \rightarrow 2 \text{KCl} + 3 \text{O}_2 \]This equation tells us that for every 2 moles of KClO_3 that decompose, 2 moles of KCl and 3 moles of O_2 are produced. The balanced equation is crucial because it enables us to determine how substances are consumed and generated.

• Reactants are the starting materials (KClO_3 in this case).
• Products are the substances formed (KCl and O_2).
• Balancing ensures conservation of mass and atoms.

Balanced equations are integral in stoichiometry as they provide the mole ratio needed to calculate unknown quantities, like the grams of reactants required to produce a desired amount of product.

Moles and Molar Mass

Understanding moles and molar mass is key to solving stoichiometry problems, such as the one involving KCl formation. The concept of a mole allows chemists to count atoms in terms that are practical for large-scale manipulation. One mole corresponds to Avogadro's number, approximately 6.022 × 10²³ entities.
The molar mass signifies the mass of one mole of a substance, expressed in g/mol. For example:

• The molar mass of KCl is about 74.55 g/mol.
• The molar mass of KClO_3 is about 122.55 g/mol.

To find the moles of a substance, divide its mass by its molar mass. For instance, to find the moles of KCl in 1.00 g:\[ \text{moles of KCl} = \frac{1.00 \text{ g}}{74.55 \text{ g/mol}} \approx 0.0134 \text{ moles} \]This step is essential as it allows conversion from mass to a usable quantity in our calculations.

Mass-to-Mass Conversion

Mass-to-mass conversion in stoichiometry is essential to bridge the gap between the known products and the required reactants. This type of problem involves interconverting masses of different substances using the stoichiometric coefficients from a balanced equation.
To solve such problems, follow these steps:

• Calculate the moles of the given substance using its molar mass.
• Use the balanced equation to find the mole ratio between the known and unknown substances.
• Convert the moles of the required substance back to mass using its molar mass.

In our equation:1. We calculated 0.0134 moles of KCl.2. Knowing the ratio from the balanced equation was 1:1 for KCl and KClO_3, we established that 0.0134 moles of KClO_3 are also needed.3. Finally, we converted these moles back to mass:\[ \text{mass of KClO}_3 = 0.0134 \text{ moles} \times 122.55 \text{ g/mol} = 1.643 \text{ g} \]Mastering these conversions allows you to solve complex stoichiometry problems efficiently, regardless of the substances used.