Final answer:
The volume of oxygen gas produced by the decomposition of 25.5 g of potassium chlorate at 2.22 atm and 25.44 degrees Celsius is 9.35 liters. This was calculated using the Ideal Gas Law after determining the moles of potassium chlorate and converting them to moles of oxygen gas.
Step-by-step explanation:
To calculate the volume of oxygen gas produced, we'll need to use the Ideal Gas Law (PV = nRT). Given that the decomposition of potassium chlorate produces oxygen gas according to the reaction 2KClO₃(s) ➡ 2KCl(s) + 3O₂(g), the first step is to determine the number of moles of potassium chlorate (KClO₃) and then use stoichiometry to find the moles of O₂ produced.
Step 1: Calculate moles of KClO₃: molecular weight of KClO₃ = 39.1 + 35.45 + 3(16) = 122.55 g/mol.
25.5 g KClO₃ × (1 mol / 122.55 g) = 0.208 mol KClO₃.
Step 2: Use stoichiometry to find moles of O₂.
(0.208 mol KClO₃) × (3 mol O₂ / 2 mol KClO₃) = 0.312 mol O₂.
Step 3: Use the Ideal Gas Law to calculate the volume of O₂.
Convert the temperature to Kelvin: T = 25.44°C + 273.15 = 298.59 K.
Now plug the values into the Ideal Gas Law formula:
V = nRT / P
V = (0.312 mol × 0.0821 L atm/mol K × 298.59 K) / 2.22 atm
V = 9.35 L
Thus, the volume of oxygen gas produced at 2.22 atm and 25.44 degrees Celsius is 9.35 liters, option C is the correct answer, and this result has three significant figures due to the given values of temperature and pressure.