Question

When heated, solid potassium chlorate \(\left(KClO_{3}\right)\) decomposes to form solid potassium chloride and oxygen gas. If 20.8 \(\mathrm{g}\) of potassium chlorate decomposes, how many liters of oxygen gas will form at STP?

Short answer

The volume of oxygen gas produced when 20.8g of potassium chlorate decomposes at STP is 5.70 liters.

Step-by-step solution

Write the balanced equation.

The balanced chemical equation for the decomposition of potassium chlorate is: \[2KClO_3 \to 2KCl + 3O_2\] This equation states that for every 2 moles of potassium chlorate that decompose, 3 moles of oxygen gas are produced.

Convert the mass of potassium chlorate to moles.

To convert the mass of potassium chlorate (20.8g) to moles, we will use its molar mass, which can be calculated as follows: Molar mass of \(KClO_3 = K + Cl + (3 \times O)\) Molar mass of potassium (K) is 39.1 g/mol, Molar mass of chloride (Cl) is 35.5 g/mol, Molar mass of oxygen (O) is 16.0 g/mol, Molar mass of \(KClO_3 = 39.1 + 35.5 + (3 \times 16.0) = 122.6 \, g/mol\) Now, we will use the molar mass to convert the mass of potassium chlorate (20.8g) to moles: Moles of \(KClO_3 = \frac{20.8\,g}{122.6\,\frac{g}{mol}} = 0.1697\,mol\)

Calculate the moles of oxygen gas produced.

Using the stoichiometry from the balanced equation, we will determine the moles of oxygen gas produced when 0.1697 moles of potassium chlorate decompose: Moles of \(O_2 = \frac{3\, moles\, of\, O_2}{2\, moles\, of\, KClO_3} \times 0.1697\, mol\) Moles of \(O_2 = 0.2546 \, mol\)

Calculate the volume of oxygen gas at STP.

At STP (standard temperature and pressure), the conditions are as follows: Temperature (T) = 273.15 K, Pressure (P) = 1 atm, Gas constant (R) = 0.0821 L atm / (mol K), Molar volume of an ideal gas at STP = 22.4 L/mol. From the molar volume of an ideal gas at STP, we have: Volume of \(O_2= 0.2546\,mol \times 22.4\,\frac{L}{mol} = 5.70\,L\) Thus, the volume of oxygen gas produced at STP is 5.70 liters.

Key concepts

Understanding Chemical Reaction Equations

Chemical reaction equations are vital components of chemistry that represent the process of converting reactants into products. In our exercise, potassium chlorate decomposes into potassium chloride and oxygen gas. The balanced chemical equation,
\[2KClO_3 \to 2KCl + 3O_2\],
reveals the mole ratio of reactants to products, essential in predicting the quantities involved in the reaction. For students, it's crucial to recognize that the coefficients indicate the proportional amounts of reactants and products. Here, 2 moles of potassium chlorate yield 2 moles of potassium chloride and 3 moles of oxygen. Understanding this ratio allows one to solve stoichiometric problems accurately. Remember to balance the equation first, as only a balanced equation represents the conservation of mass in chemical reactions, a fundamental law in chemistry.

Mole Conversions Simplified

Mole conversions are at the heart of stoichiometry. They allow chemists to count particles by weighing them. Every substance has a unique molar mass, which can be used to convert between grams and moles. In our problem:
\[Moles\text{ of }KClO_3 = \frac{20.8\,g}{122.6\,\frac{g}{mol}} = 0.1697\,mol\],
we used the molar mass of potassium chlorate to convert the given mass to moles. It's important for students to correctly calculate the molar mass by adding up the atomic masses of each element in the compound. Once conversion to moles is done, these moles are then used as an input to understand the chemical equation's mole ratio, thus bridging the gap between the mass of a substance and the number of particles contained in that mass.

Gas Laws at Work

Gas laws describe how gases behave in different conditions of temperature, pressure, and volume. A crucial point for students to grasp is that at standard temperature and pressure (STP), one mole of an ideal gas occupies 22.4 liters. This makes calculations at STP straightforward:
\[Volume \text{ of } O_2 = 0.2546\,mol \times 22.4\,\frac{L}{mol} = 5.70\,L\].
By knowing the amount of gas in moles, we directly determine its volume. Remember, STP refers to a temperature of 273.15 K and a pressure of 1 atm. Deviations from STP require the application of other gas laws (like Boyle's or Charles's law), but for this exercise, the molar volume concept is sufficient. Encouraging students to always check the conditions under which gas behavior is considered helps avoid confusion and errors in calculations.