Calculate the volume of oxygen gas produced at 31.65 atm and 55.78 oC by the complete decomposition of 30.23 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 find the number of moles of potassium chlorate (KClO3). The molar mass of KClO3 is approximately 122.55 g/mol. So, the number of moles of KClO3 is 30.23 g / 122.55 g/mol = 0.2468 mol.

2. From the balanced chemical equation, we know that 2 moles of KClO3 produce 3 moles of O2. Therefore, 0.2468 mol of KClO3 will produce 0.2468 mol * (3/2) = 0.3702 mol 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, so T = 55.78 + 273.15 = 328.93 K.

4. The value of R depends on the units used. Since the pressure is given in atm, we will use R = 0.0821 L·atm/(K·mol). So, we can rearrange the ideal gas law to solve for V: V = nRT / P = 0.3702 mol * 0.0821 L·atm/(K·mol) * 328.93 K / 31.65 atm = 0.308 L or 308 mL.

So, the volume of oxygen gas produced by the complete decomposition of 30.23 g of potassium chlorate at 31.65 atm and 55.78 oC is approximately 308 mL.