Question

A chemist performs the following reaction: \(2 \mathrm{KClO}_{3}(\mathrm{~s}) \rightarrow 2 \mathrm{KCl}(\mathrm{s})+3 \mathrm{O}_{2}(\mathrm{~g}) .\) He collects the \(\mathrm{O}_{2}\) gas by water displacement at \(20^{\circ} \mathrm{C}\). He observes a pressure of 753 \(\mathrm{mm} \mathrm{Hg}\). Assuming the pressure of water vapor is \(17.5 \mathrm{~mm}\) \(\mathrm{Hg}\) at \(20^{\circ} \mathrm{C}\) and he started with \(1.28 \mathrm{~g}\) of potassium chlorate \(\left(\mathrm{KClO}_{3}\right)\) what volume of gas is produced? \((\mathrm{R}=.0821\) \(\ell\) -atm/mole \({ }^{\circ} \mathrm{K}\).)

Short answer

The volume of the O2 gas produced in the given reaction is 0.320 L.

Step-by-step solution

Calculate the molar mass of KClO3

First, we need to find the molar mass of KClO3 by summing the molar masses of its constituent elements: K (39.10 g/mol), Cl (35.45 g/mol), and O (3 * 16.00 g/mol). The final result is \(M_{KClO_3} = 122.55 g/mol\).

Calculate the moles of KClO3

Now, we can determine the moles of KClO3 by dividing the given mass by the molar mass: \[n_{KClO_3} = \frac{1.28g}{122.55 g/mol} = 0.0104 \, mol\] #Step 2: Find moles of O2 produced#

Determine the stoichiometry ratio

From the balanced chemical equation, we can see that 2 moles of KClO3 produce 3 moles of O2. Therefore, the stoichiometry ratio is 3:2.

Calculate the moles of O2 produced

Now, we can find the moles of O2 produced by multiplying the moles of KClO3 by the stoichiometry ratio: \[n_{O_2} = n_{KClO_3} \times \frac{3 \, mol \, O_2}{2\, mol\, KClO_3} = 0.0104\, mol \times \frac{3}{2} = 0.0156 \, mol\] #Step 3: Find the volume of O2 gas produced#

Calculate the partial pressure of O2

We are given the total pressure observed (753 mm Hg) and the pressure of water vapor at 20°C (17.5 mm Hg). We can find the partial pressure of O2 by subtracting the pressure of the water vapor: \[P_{O_2} = 753 - 17.5 = 735.5 \, mm\, Hg\] Convert this pressure to atmospheres by dividing by 760: \[P_{O_2} = \frac{735.5\, mm\, Hg}{760\, mm\, Hg/atm} = 0.9674\, atm\]

Use the Ideal Gas Law to find the volume

We can now use the Ideal Gas Law to find the volume of the O2 gas produced. The temperature in Kelvin is 20°C+273.15 = 293.15 K, and the ideal gas constant R is 0.0821 L·atm/(mol·K). The Ideal Gas Law equation is: \[PV = nRT\] Solving for the volume: \[V = \frac{nRT}{P} = \frac{(0.0156\, mol)(0.0821\, L\, atm/(mol\, K))(293.15\, K)}{0.9674\, atm} = 0.320\, L\] #Conclusion# The volume of the O2 gas produced is 0.320 L.

Key concepts

Molar Mass Calculation

Understanding the molar mass of a substance is a crucial step in many chemical reaction calculations. In essence, the molar mass corresponds to the weight of one mole (6.022\( \times \)10\(^{23}\) particles) of a given substance and is expressed in grams per mole (g/mol). The process to determine this value involves summing up the atomic masses of each element that makes up the compound, based on its chemical formula.For example, potassium chlorate (\(KClO_3\)) is composed of one potassium (K) atom, one chlorine (Cl) atom, and three oxygen (O) atoms. The molar mass is calculated by adding the atomic mass of K (39.10 g/mol), Cl (35.45 g/mol), and O (16.00 g/mol multiplied by 3 because there are three oxygen atoms), resulting in a molar mass of 122.55 g/mol for \(KClO_3\).Knowing the molar mass allows us to convert between the mass of a substance and the number of moles, which is indispensable for stoichiometry calculations.

Stoichiometry

Stoichiometry is the quantitative aspect of chemistry that involves calculating the amounts of reactants and products in a chemical reaction. It relies heavily on the Law of Conservation of Mass, which states that mass is neither created nor destroyed in a chemical reaction. Stoichiometry uses the balanced chemical equation as a starting point, which provides the ratio of the number of moles of each reactant and product involved in the reaction.In the given exercise, we determined the stoichiometry ratio from the balanced equation \(2KClO_3(s) \rightarrow 2KCl(s) + 3O_2(g)\). This tells us that 2 moles of \(KClO_3\) produce 3 moles of \(O_2\). When we calculated the moles of \(KClO_3\), this ratio allowed us to find the corresponding moles of \(O_2\) produced. It’s essentially a proportion problem where the coefficients in the chemical equation give the relative amounts of reactants and products involved.

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry that relates the pressure, volume, temperature, and number of moles of an ideal gas. The law is usually stated as \(PV = nRT\), where P is the pressure, V is the volume, n is the number of moles, R is the universal gas constant, and T is the temperature in Kelvin.This equation assumes that gas particles do not interact and there are no intermolecular forces affecting them, which is generally a good approximation under standard conditions. When solving for one of these variables, the others must be known or calculable.In the exercise, to find the volume of oxygen gas produced, we first subtracted the water vapor pressure from the observed pressure to get the partial pressure of \(O_2\). Then, using the Ideal Gas Law and solving for V, we found the volume while taking into account the temperature in Kelvin and utilizing the given value for R. The result is a practical application of how the Ideal Gas Law can predict the behavior of gases under specific conditions.