Problem 79 - Chemistry and Chemical Reactivity

Question

You are given \(1.56 \mathrm{g}\) of a mixture of \(\mathrm{KClO}_{3}\) and \(\mathrm{KCl}\). When heated, the KClOg decomposes to KCl and \(\mathrm{O}_{2}\), $$ 2 \mathrm{KClO}_{3}(\mathrm{s}) \rightarrow 2 \mathrm{KCl}(\mathrm{s})+3 \mathrm{O}_{2}(\mathrm{g}) $$ and \(327 \mathrm{mL}\) of \(\mathrm{O}_{2}\) with a pressure of \(735 \mathrm{mm}\) Hg is collected at \(19^{\circ} \mathrm{C}\). What is the weight percentage of \(\mathrm{KClO}_{3}\) in the sample?

Step-by-Step Solution

Answer

The weight percentage of \(\text{KClO}_3\) in the sample is 68.08\%.

Step 1: Understand the Chemical Reaction

The decomposition reaction is given as \(2 \text{KClO}_3 \rightarrow 2 \text{KCl} + 3 \text{O}_2\). This tells us that 2 moles of KClO3 produce 3 moles of O2. We will use this stoichiometric ratio to find the amount of KClO3.

Step 2: Use the Ideal Gas Law

We need to first find the number of moles of \(\text{O}_2\). We use the ideal gas equation, \(PV = nRT\), where \(P = 735 \text{ mm Hg} = 0.967\text{ atm}\), \(V = 327 \text{ mL} = 0.327 \text{ L}\), \(R = 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1}\), and \(T = 19^\circ C = 292 \text{ K}\). Solve for \(n\):

Step 3: Solve for Moles of Oxygen

\[n = \frac{PV}{RT} = \frac{0.967 \times 0.327}{0.0821 \times 292} = 0.0130 \text{ mol of } \text{O}_2\]

Step 4: Calculate Moles of KClO3

Using the stoichiometric ratio from the equation, \(2 \text{ mol KClO}_3 : 3 \text{ mol O}_2\), we find the moles of KClO3: \[n_{\text{KClO}_3} = \frac{2}{3} \times 0.0130 = 0.00867 \text{ mol KClO}_3\]

Step 5: Convert Moles to Grams

Now we convert moles of KClO3 to grams using its molar mass (KClO3 = 122.55 g/mol): \[\text{mass of KClO}_3 = 0.00867 \text{ mol} \times 122.55 \text{ g/mol} = 1.062 \text{ g}\]

Step 6: Calculate Weight Percentage

The total mass of the mixture is 1.56 g. The percentage of KClO3 is given by: \[\text{Weight percentage of KClO}_3 = \left(\frac{1.062}{1.56}\right) \times 100\% = 68.08\%\]

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation connecting pressure, volume, temperature, and the number of moles of a gas. It's expressed as \(PV = nRT\), where:

• \(P\) is pressure, generally measured in atmospheres (atm).
• \(V\) is volume, typically in liters (L).
• \(n\) is the number of moles of the gas.
• \(R\) is the universal gas constant, \(0.0821\) L atm K\(^{-1}\) mol\(^{-1}\).
• \(T\) is the temperature in Kelvin (K).

In the given exercise, the pressure is converted from mm Hg to atm because the constant \(R\) is in terms of atm. Similarly, milliliters must be converted to liters to maintain consistent units. By inserting these values, we can solve for \(n\), the number of moles of \(\mathrm{O}_2\) gas produced.

Chemical Reactions

Chemical reactions involve rearranging atoms to transform initial substances, known as reactants, into new substances, called products. In this case, the reaction is the decomposition of \(\text{KClO}_3\) (potassium chlorate) into \(\text{KCl}\) (potassium chloride) and \(\text{O}_2\) (oxygen gas). The balanced equation for the reaction is: \[2 \text{KClO}_3 \rightarrow 2 \text{KCl} + 3 \text{O}_2\]Balanced reactions mean the same number of each type of atom appears on both sides of the equation. This stoichiometric balance helps us calculate the amounts of reactants and products involved, by maintaining consistent proportions. Understanding the stoichiometry allows us to infer the production of \(\text{O}_2\) directly affects how much \(\text{KClO}_3\) was initially present in the mixture.

Mole Concept

The mole concept is a cornerstone of chemistry used to express amounts of a chemical substance. One mole corresponds to Avogadro's number (\(6.022 \times 10^{23}\)), which is the number of atoms, ions, or molecules contained in one mole of a substance.

• This exercise set up uses stoichiometry to relate moles of \(\text{O}_2\) to moles of \(\text{KClO}_3\).
• From our reaction, the stoichiometry indicates 2 moles of \(\text{KClO}_3\) produce 3 moles of \(\text{O}_2\).
• With the moles of \(\text{O}_2\) known, the moles of \(\text{KClO}_3\) can be calculated using the ratio \(\frac{2}{3}\).

In the example, knowing the moles of \(\text{O}_2\) allows for the backward calculation to find the original moles of \(\text{KClO}_3\) that decomposed to produce that particular amount.

Weight Percentage Calculation

Determining the weight percentage of a substance within a mixture involves comparing the mass of the specific substance to the total mass of the mixture.

Steps for Calculation:

• First, calculate the mass of the interested compound, \(\text{KClO}_3\), using its mole calculation and molar mass.
• Weight percentage is then found using the formula \(\left(\frac{\text{mass of component}}{\text{total mass of mixture}}\right) \times 100\%\).

In this problem, once the moles of \(\text{KClO}_3\) were determined, they were converted to grams using its molar mass (\(122.55 \text{ g/mol}\)). This mass was then divided by the total mixture mass, and the result was multiplied by 100 to get the percentage. For this setup, the calculations revealed \(\text{KClO}_3\) formed approximately \(68.08\%\) of the mass of the mixture.