Question

Potassium chlorate decomposes upon heating as follows:

2KClO3(s) → 2KCl(s) + 3O2(g)
A 2.72-g sample of KClO3 decomposes, and the oxygen at 23.4°C and 0.935 atm is collected. What volume of oxygen gas will be collected, assuming 100% yield?

Answer 1

To find the volume of oxygen gas collected, we need to use the Ideal Gas Law equation and the given pressure and temperature. By converting the temperature to Kelvin and the pressure to kilopascals, we can use the formula PV = nRT to solve for the volume. We also need to calculate the number of moles of oxygen gas produced using the mass of KClO3 decomposed and the molar mass of KClO3.

Step-by-step explanation:

To find the volume of oxygen gas collected, we need to use the Ideal Gas Law equation: PV = nRT. In this equation, P represents pressure, V represents volume, n represents the number of moles, R is the ideal gas constant, and T represents temperature. We are given the pressure (0.935 atm) and temperature (23.4°C), and we need to find the volume of oxygen gas.

First, we need to convert the temperature to Kelvin by adding 273.15 to the Celsius temperature: 23.4 + 273.15 = 296.55 K. Next, we need to convert the pressure from atm to kPa by multiplying it by 101.325: 0.935 atm * 101.325 kPa/atm = 94.74 kPa. Now we can rearrange the Ideal Gas Law equation to solve for volume:

V = nRT / P

We need to find the number of moles of oxygen gas to use in the equation. From the balanced chemical equation, we know that 2 moles of KClO3 produce 3 moles of O2. So, we can calculate the number of moles of O2 produced by dividing the mass of KClO3 decomposed by the molar mass of KClO3 and multiplying by 3/2:

Moles of O2 = (2.72 g / 122.55 g/mol) * (3/2) = 0.0669 mol O2

Finally, we can plug the values into the equation:

V = (0.0669 mol) * (0.0821 L·atm/mol·K) * (296.55 K) / (94.74 kPa) = 0.202 L