Problem 81 - 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 KClOs decomposes to KCl and \(\mathrm{O}_{2}\), $$ 2 \mathrm{KClO}_{3}(\mathrm{s}) \longrightarrow 2 \mathrm{KCl}(\mathrm{s})+3 \mathrm{O}_{2}(\mathrm{g}) $$ and \(327 \mathrm{mL}\) of \(\mathrm{O}_{2}\) with a pressure of \(735 \mathrm{mm} \mathrm{Hg}\) 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 \( \mathrm{KClO}_{3} \) in the sample is 70.13\%.

Step 1: Convert Pressure to Atmospheres

First, we need to convert the pressure from mmHg to atm to use in the ideal gas law. \[ 1 \text{ atm} = 760 \text{ mmHg} \] Thus,\[ P = \frac{735 \text{ mmHg}}{760 \text{ mmHg/atm}} = 0.9671 \text{ atm} \]

Step 2: Convert Temperature to Kelvin

Convert the temperature from Celsius to Kelvin using the formula:\[ T_{K} = T_{C} + 273.15 \]Thus,\[ T_{K} = 19 + 273.15 = 292.15 \text{ K} \]

Step 3: Use the Ideal Gas Law

Apply the ideal gas law to find the number of moles of \( \mathrm{O}_{2} \). The ideal gas law is given by\[ PV = nRT \]Where \( R = 0.0821 \text{ L atm/mol K} \).First, convert volume to liters:\[ V = 327 \text{ mL} = 0.327 \text{ L} \]Rearrange to solve for \( n \):\[ n = \frac{PV}{RT} = \frac{0.9671 \times 0.327}{0.0821 \times 292.15} = 0.0134 \text{ moles of } \mathrm{O}_{2} \]

Step 4: Find Moles of KClO3 Decomposed

Use stoichiometry from the balanced chemical equation:\[ 2 \mathrm{KClO}_{3} \rightarrow 2 \mathrm{KCl} + 3 \mathrm{O}_{2} \]For 3 moles of \( \mathrm{O}_{2} \), 2 moles of \( \mathrm{KClO}_{3} \) are required.Calculate moles of \( \mathrm{KClO}_{3} \):\[ \text{Moles of } \mathrm{KClO}_{3} = \frac{2}{3} \times 0.0134 \approx 0.00893 \text{ moles} \]

Step 5: Calculate Mass of KClO3

Calculate the mass of \( \mathrm{KClO}_{3} \) using its molar mass (\( K = 39.10, Cl = 35.45, O = 16.00 \)).\[ \text{Molar mass of } \mathrm{KClO}_{3} = 39.10 + 35.45 + 3 \times 16.00 = 122.55 \text{ g/mol} \]Mass of \( \mathrm{KClO}_{3} \):\[ \text{Mass} = 0.00893 \times 122.55 = 1.094 \text{ g} \]

Step 6: Calculate Weight Percentage of KClO3

Finally, calculate the weight percentage of \( \mathrm{KClO}_{3} \) in the sample.\[ \text{Weight percentage} = \frac{\text{mass of } \mathrm{KClO}_{3}}{\text{total mass}} \times 100\%\]\[ = \frac{1.094}{1.56} \times 100\%\approx 70.13\% \]

Key Concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that involves the calculation of reactants and products in chemical reactions. It is based on the balanced chemical equation which provides the proportions of reactants and products. In the given exercise, stoichiometry helps us determine the relationship between the decomposed potassium chlorate (\(\mathrm{KClO}_3\)) and the oxygen gas (\(\mathrm{O}_2\)) produced.
From the balanced chemical equation \(2 \mathrm{KClO}_{3} \rightarrow 2 \mathrm{KCl} + 3 \mathrm{O}_{2}\), we know that two moles of \(\mathrm{KClO}_3\) produce three moles of \(\mathrm{O}_2\). Thus, if we have the amount of \(\mathrm{O}_2\) produced, we can find out how much \(\mathrm{KClO}_3\) was initially present.

• Firstly, calculate the moles of \(\mathrm{O}_2\) using the ideal gas law.
• Then, use the relationship from the equation: \(\mathrm{Moles\ of\ } KClO_3 = \frac{2}{3} \times \mathrm{Moles\ of\ } O_2\).

This proportion helps in translating the quantity of one substance to another, adhering to the law of conservation of mass.

Molar Mass Calculation

Calculating molar mass is essential for understanding the composition of compounds in a chemical reaction. Molar mass, defined as the mass of one mole of a substance, is a critical piece of the puzzle in the given exercise.
To calculate the molar mass of \(\mathrm{KClO}_3\), we must know the individual atomic masses of potassium (\(K\)), chlorine (\(Cl\)), and oxygen (\(O\)).

• Potassium (\(K\)) has an atomic mass of 39.10 g/mol.
• Chlorine (\(Cl\)) has an atomic mass of 35.45 g/mol.
• Oxygen (\(O\)) has an atomic mass of 16.00 g/mol, and since there are three oxygen atoms in \(\mathrm{KClO}_3\), we multiply by 3.

Adding these together, the molar mass of \(\mathrm{KClO}_3\) is calculated as:\[\text{Molar Mass of } \mathrm{KClO}_3 = 39.10 + 35.45 + 3 \times 16.00 = 122.55 \text{ g/mol}\]This value is crucial for converting moles of \(\mathrm{KClO}_3\) to its mass, which is needed to find the weight percentage of \(\mathrm{KClO}_3\) in the sample.

Gas Pressure Conversion

In gas law problems, it's crucial to work with consistent units, specially when using the ideal gas law. Often, the pressure of a gas might be given in units like mmHg or torr, but for calculations, it's usually necessary to convert these to atmospheres (atm).
In the original exercise, the pressure of \(\mathrm{O}_2\) collected was given in mmHg and required conversion.

• The conversion factor is: \(1 \text{ atm} = 760 \text{ mmHg}\).
• For the pressure of 735 mmHg, the conversion to atm is computed as follows:\[P = \frac{735 \text{ mmHg}}{760 \text{ mmHg/atm}} = 0.9671 \text{ atm}\]

Converting the pressure correctly is mandatory for the accurate application of the ideal gas law \( (PV = nRT) \) when determining the number of moles of gas. Understanding this step ensures that all components of the formula are in the proper units, leading to valid and reliable results.