Calculate the volume of oxygen gas produced at 69.97 atm and 24.36 oC by the complete decomposition of 96.2 g of potassium chlorate. The balanced equation for this reaction is …2KClO3 ⟶ 2KCl + 3O2

To solve this problem, we need to use the ideal gas law and stoichiometry. Here are the steps:

1. First, we need to convert the mass of potassium chlorate (KClO3) to moles. The molar mass of KClO3 is approximately 122.55 g/mol. So, 96.2 g of KClO3 is 96.2 g / 122.55 g/mol = 0.785 moles.

2. From the balanced chemical equation, we know that 2 moles of KClO3 produce 3 moles of O2. Therefore, 0.785 moles of KClO3 will produce 0.785 * (3/2) = 1.178 moles of O2.

3. Now we can use the ideal gas law to find the volume of O2. The ideal gas law is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin. We need to convert the temperature from Celsius to Kelvin by adding 273.15 to it. So, T = 24.36°C + 273.15 = 297.51 K.

4. We are looking for the volume, so we rearrange the ideal gas law to solve for V: V = nRT / P. We use the value of R that has the same units as our pressure, which is 0.0821 L·atm/(K·mol). So, V = 1.178 moles * 0.0821 L·atm/(K·mol) * 297.51 K / 69.97 atm = 3.96 L.

So, the volume of oxygen gas produced by the complete decomposition of 96.2 g of potassium chlorate at 69.97 atm and 24.36°C is approximately 3.96 liters.