Using the Ideal Gas Law, we determined that the volume of oxygen produced by the catalytic decomposition of 25.5 g potassium chlorate is 4.31 liters.
To calculate the volume of oxygen produced by the catalytic decomposition of potassium chlorate, we can use the Ideal Gas Law equation: 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.
First, we need to determine the number of moles of oxygen produced. We can use the molar mass of potassium chlorate (KClO3) and the given mass of potassium chlorate (25.5 g) to find the number of moles.
1. Calculate the molar mass of potassium chlorate (KClO3):
K: 1 atom × 39.10 g/mol = 39.10 g/mol
Cl: 1 atom × 35.45 g/mol = 35.45 g/mol
O: 3 atoms × 16.00 g/mol = 48.00 g/mol
Total molar mass = 39.10 g/mol + 35.45 g/mol + 48.00 g/mol = 122.55 g/mol
2. Calculate the number of moles of potassium chlorate:
Moles = Mass / Molar mass
Moles = 25.5 g / 122.55 g/mol = 0.208 mol
Now, we can calculate the volume of oxygen using the Ideal Gas Law equation:
PV = nRT
3. Convert the given temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 25.44°C + 273.15 = 298.59 K
4. Rearrange the Ideal Gas Law equation to solve for V:
V = (nRT) / P
V = (0.208 mol × 0.0821 L atm/mol K × 298.59 K) / 2.22 atm
V = 4.31 L
Therefore, the volume of oxygen produced by the catalytic decomposition of 25.5 g potassium chlorate is 4.31 liters.