Question

A 3.57 g sample of a \(\mathrm{KCl}-\mathrm{KClO}_{3}\) mixture is decomposed by heating and produces \(119 \mathrm{mL} \mathrm{O}_{2}(\mathrm{g})\) measured at \(22.4^{\circ} \mathrm{C}\) and \(738 \mathrm{mm} \mathrm{Hg}\). What is the mass percent of \(\mathrm{KClO}_{3}\) in the mixture? $$2 \mathrm{KClO}_{3}(\mathrm{s}) \longrightarrow 2 \mathrm{KCl}(\mathrm{s})+3 \mathrm{O}_{2}(\mathrm{g})$$

Short answer

The mass percent of \(KClO_3\) in the mixture is 11.1%.

Step-by-step solution

Convert given conditions to standard form

First, it's important to remember that the volume should be in liters and the pressure should be in atm for the ideal gas laws equation. The given volume is 119 mL, which can be converted to liters by dividing by 1000 (119 mL = 0.119 L), and the given pressure is 738 mm Hg, which can be converted to atmospheres by dividing by 760 (738 mmHg = 0.971 atm). The temperature is given in degree celsius and it needs to be converted to Kelvins (K = degree celsius + 273). Thus, the temperature is 22.4 + 273 = 295.4 K.

Use ideal gas law to find moles of O2

Use the ideal gas equation \(PV = nRT\), where P is the pressure (0.971 atm), V is the volume (0.119 L), T is the temperature (295.4 K), n is the number of moles (which we are trying to find), and R is the ideal gas constant (0.0821 L·atm/mol·K). Solve the equation for n: \(n = \frac{PV}{RT} = \frac{(0.971 atm)(0.119 L)}{(0.0821 L⋅atm/mol⋅K)(295.4 K)} = 0.00484 mol\) of \(O_2\).

Use stoichiometry to get moles of KClO3

We know that from the balanced chemical equation provided, 2 moles of \(KClO_3\) yields 3 moles of \(O_2\). Thus, converting the moles of \(O_2\) to moles of \(KClO3\), we get \(n_{KClO3} = 0.00484 mol O2 * 2/3 = 0.00323 mol of KClO3\).

Calculate mass of KClO3

Multiply the obtained moles of \(KClO_3\) by the molar mass of \(KClO_3\) (122.55 g/mol) to get the mass: \(mass_{KClO3} = 0.00323 mol * 122.55 g/mol = 0.396 g\)

Calculate mass percent of KClO3

Finally, apply the mass percent formula, \(mass percent = \frac{mass of part}{mass of whole} \times 100%\). In this context, it will be \(mass percent_{KClO3} = \frac{0.396 g}{3.57 g} \times 100% = 11.1%\)

Key concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry. It helps us understand the quantitative relationships between reactants and products in a chemical reaction. When dealing with stoichiometry:

• We rely on balanced chemical equations to provide a mole-to-mole ratio between substances.
• A balanced equation shows the exact amounts (in moles) of substances that react and form products.
• By using these relationships, we can calculate how much of a product forms from a given amount of reactant or vice versa.

In the exercise, we use the stoichiometric ratio from the given chemical equation for the decomposition of potassium chlorate (\(2KClO_3 \rightarrow 2KCl + 3O_2\)). This tells us that two moles of \(KClO_3\) produce three moles of \(O_2\). By determining the moles of \(O_2\) formed, we back-calculate to find the moles of \(KClO_3\) that decomposed. This illustrates stoichiometry's power: converting measurable gas volume to a specific mass percentage in the mixture.

Ideal Gas Law

The Ideal Gas Law is a useful equation (\(PV = nRT\)) that relates the physical properties of gases. In this law:

• \(P\) stands for pressure, \(V\) for volume, \(n\) for the number of moles, \(R\) is the gas constant, and \(T\) is the temperature.
• It allows us to switch between the physical properties of a gas, and calculates moles when the other quantities are known.

In our problem, we start with a provided volume of \(O_2\) given at certain experimental conditions (not standard conditions). First, it's necessary to convert the units to apply the ideal gas law formulas correctly:

• Pressure in mm Hg is converted to atm by dividing by 760.
• Volume from mL to L by dividing by 1000.
• Temperature from Celsius to Kelvin by adding 273.15.

With these adjusted figures, the Ideal Gas Law formula calculates the moles of \(O_2\) gas. This number helps us further to apply the stoichiometric concepts, establishing links between experimental conditions and theoretical calculations.

KClO3 Decomposition

The process of \(KClO_3\) decomposition is a classic chemical reaction involving heat. In this reaction:

• \(KClO_3\) is heated to break down into \(KCl\) and \(O_2\) gas.
• The reaction is represented by the balanced equation: \(2KClO_3(s) \rightarrow 2KCl(s) + 3O_2(g)\).
• This tells us that two moles of potassium chlorate decompose to produce two moles of potassium chloride and three moles of oxygen gas.

Understanding this decomposition allows chemists to predict and calculate the amount of gas (in this case, oxygen) expected under certain conditions. Through heating, the production of oxygen gas can be measured, and using stoichiometry, the amount of initial \(KClO_3\) in a mixture can be deduced. This forms the crux of calculating the mass percent of \(KClO_3\) in the original mixture, connecting theoretical chemical principles with practical laboratory outcomes.