Oxygen gas can be generated by heating KClO3 in the presence of some MnO2.
2KClO3 → 2KCl + 3O2
If all the KClO3 in a mixture containing 1.00 g KClO3 and 0.25 g MnO2 is decomposed, how many liters of O2 gas will be generated at T = 24.0°C and P = 755.6 torr?
Convert 1.0 g KClO3 to mols. mols = grams/molar mass. (I am assuming the MnO2 is a catalyst and does not enter the reaction).
Using the coefficients in the balanced equation convert mols KClO3 to mols O2.
Then use PV = nRT and the conditions listed in the problem to solve for V. Remember to convert T to kelvin.
To find the number of liters of O2 gas generated, we need to use the ideal gas law equation:
PV = nRT
Where:
P = Pressure in atmospheres
V = Volume in liters
n = Number of moles
R = Ideal gas constant
T = Temperature in Kelvin
Given:
T = 24.0°C = 24.0 + 273.15 = 297.15 K
P = 755.6 torr = 755.6/760 = 0.9947 atm
Molar mass of KClO3 = 122.55 g/mol
Molar mass of O2 = 32.0 g/mol
Step 1: Calculate the number of moles of KClO3
Moles of KClO3 = Mass of KClO3 / Molar mass of KClO3
Moles of KClO3 = 1.00 g / 122.55 g/mol = 0.00816 mol
Step 2: Calculate the number of moles of O2 produced
According to the balanced equation, 2 moles of KClO3 produce 3 moles of O2.
So, moles of O2 = (0.00816 mol KClO3) x (3 mol O2 / 2 mol KClO3) = 0.01224 mol
Step 3: Calculate the volume of O2 at the given conditions
Convert temperature from Celsius to Kelvin
T = 297.15 K
Now, rearrange the ideal gas law equation to solve for volume:
V = (nRT) / P
Substitute the values into the equation:
V = (0.01224 mol) x (0.0821 L·atm/(mol·K)) x (297.15 K) / 0.9947 atm
V ≈ 0.351 L
Therefore, at a temperature of 24.0°C and pressure of 755.6 torr, approximately 0.351 liters of O2 gas will be generated when all the KClO3 in the mixture is decomposed.