Problem 87 - Chemistry and Chemical Reactivity

Question

You are given 1.56 g of a mixture of \(\mathrm{KClO}_{3}\) and KCl. When heated, the \(\mathrm{KClO}_{3}\) decomposes to KCl and \(\mathbf{O}_{2}\) $$ 2 \mathrm{KClO}_{3}(\mathrm{s}) \rightarrow 2 \mathrm{KCl}(\mathrm{s})+3 \mathrm{O}_{2}(\mathrm{g}) $$ and \(327 \mathrm{mL}\) of \(\mathrm{O}_{2}\) with a pressure of \(735 \mathrm{mm} \mathrm{Hg}\) 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 approximately 35.4%.

Step 1: Use Ideal Gas Law to Find Moles of O2

First, we need to determine the moles of \(\mathrm{O}_{2}\) generated during the decomposition. Use the ideal gas law: \(pV = nRT\).Given: - Pressure (\(p\)) = 735 mmHg = 0.967 atm (since 1 atm = 760 mmHg) - Volume (\(V\)) = 327 mL = 0.327 L - Temperature (\(T\)) = 19°C = 292 K (since \(T(\text{K}) = T(\text{°C}) + 273.15\)) - \(R\) (ideal gas constant) = 0.0821 L atm/mol KPlug these values into the ideal gas law:\[n = \frac{pV}{RT} = \frac{(0.967\, \text{atm}) \times (0.327\, \text{L})}{0.0821\, \text{L atm/mol K} \times 292\, \text{K}}\]Calculate \(n\).

Step 2: Calculate Moles of KClO3

According to the balanced equation, 2 moles of \(\mathrm{KClO}_{3}\) produce 3 moles of \(\mathrm{O}_{2}\). Use this stoichiometric relationship to find moles of \(\mathrm{KClO}_{3}\):\[\frac{2 \text{ moles } \mathrm{KClO}_{3}}{3 \text{ moles } \mathrm{O}_{2}} = \frac{x \text{ moles } \mathrm{KClO}_{3}}{\text{calculated moles of } \mathrm{O}_{2}}\]Solve for \(x\), which represents the moles of \(\mathrm{KClO}_{3}\).

Step 3: Convert Moles of KClO3 to Grams

Now that we have moles of \(\mathrm{KClO}_{3}\), convert it to grams using its molar mass:- Molar mass of \(\mathrm{KClO}_{3}\) = 122.55 g/mol\[\text{grams of } \mathrm{KClO}_{3} = x \text{ moles } \times 122.55 \text{ g/mol}\]

Step 4: Calculate Weight Percentage of KClO3

Finally, calculate the weight percentage of \(\mathrm{KClO}_{3}\) in the sample:\[\text{Weight percent} = \left(\frac{\text{grams of } \mathrm{KClO}_{3}}{1.56\, \text{g}}\right) \times 100\]%Compute this value to find the percentage of \(\mathrm{KClO}_{3}\).

Key Concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that refers to the calculation of reactants and products in chemical reactions. It is essential for determining how much of each reactant is needed to produce a desired amount of product. In the context of our exercise, stoichiometry helps to understand the relationship between the decomposition of \(\mathrm{KClO}_{3}\) and the formation of \(\mathrm{O}_{2}\). According to the balanced chemical equation:

\[2 \mathrm{KClO}_{3}(\mathrm{s}) \rightarrow 2 \mathrm{KCl}(\mathrm{s})+3 \mathrm{O}_{2}(\mathrm{g})\]

- Two moles of \(\mathrm{KClO}_{3}\) decompose to produce two moles of \(\mathrm{KCl}\) and three moles of \(\mathrm{O}_{2}\). - This means, for every mole of \(\mathrm{KClO}_{3}\), 1.5 moles of \(\mathrm{O}_{2}\) are produced. - This stoichiometric ratio is crucial for calculating how much \(\mathrm{KClO}_{3}\) was present based on the amount of \(\mathrm{O}_{2}\) collected.

Understanding stoichiometry allows us to convert between moles of reactants and products using their respective coefficients from the balanced equation. It ensures that mass is conserved and provides the basis for further calculations, such as determining molar masses or converting moles to grams.

Weight Percentage Calculation

Weight percentage is a way to express the composition of a particular compound within a mixture. It indicates the concentration of a component as a percentage of the total mixture's mass. In the exercise, we aim to find the weight percentage of \(\mathrm{KClO}_{3}\) in a given sample.

- First, calculate the grams of \(\mathrm{KClO}_{3}\) present using the moles obtained from the stoichiometry step and the molar mass of \(\mathrm{KClO}_{3}\), which is 122.55 g/mol. - Once you have the mass of \(\mathrm{KClO}_{3}\) in grams, divide it by the total mass of the sample (1.56 g in this case). - Multiply this number by 100 to convert it into a percentage.

This calculation gives you the proportion of \(\mathrm{KClO}_{3}\) relative to the entire mixture. Weight percentage is a useful and straightforward way to quantify the composition of a sample, especially in analyses where only partial decomposition or reactions occur in mixtures.

Gas Collection

Gas collection is a technique used to measure the volume of gas produced or consumed in a chemical reaction. In this specific exercise, the oxygen gas produced from the decomposition of \(\mathrm{KClO}_{3}\) was measured in a controlled environment.

To accurately determine the volume of the gas collected:

• Make sure the conditions (temperature and pressure) are known and controlled. In this exercise, \(\mathrm{O}_{2}\) was collected at a pressure of 735 mmHg and a temperature of 19°C.
• Use the volume collected, together with the Ideal Gas Law equation \(pV = nRT\), to find the moles of gas.

This method allows chemists to experimentally quantify gases even in small amounts. Moreover, understanding the collection conditions enables more precise calculations of the gas's properties, supporting the stoichiometric analysis.