Potassium perchlorate decomposes as follows: KClO4 (s) --> KCl (s) + 2 O2 (g)
If 269 mL of wet O2 (g) is collected over water at 25 c (vapor pressure= 23.8 torr) and the barometric pressure is 748.0 torr, how many grams of potassium perchlorate have decomposed?
See 10:14pm post.
To find the mass of potassium perchlorate that decomposed, we need to use the ideal gas law to calculate the amount of O2 gas produced and then use stoichiometry to relate it to the mass of potassium perchlorate.
First, let's calculate the partial pressure of the dry O2 gas. The total pressure is the sum of the vapor pressure of water and the partial pressure of the dry O2 gas:
Total Pressure = Vapor Pressure of Water + Partial Pressure of O2 gas
Total Pressure = 748.0 torr (given)
Vapor Pressure of Water = 23.8 torr (given)
Partial Pressure of O2 gas = Total Pressure - Vapor Pressure of Water
Partial Pressure of O2 gas = 748.0 torr - 23.8 torr
Partial Pressure of O2 gas = 724.2 torr
Now, let's convert the volume of the collected O2 gas from milliliters to liters:
Volume of O2 gas (L) = 269 mL * (1 L / 1000 mL)
Volume of O2 gas (L) = 0.269 L
We can now use the ideal gas law to calculate the moles of O2 gas:
PV = nRT
Where:
P = Partial Pressure of O2 gas (in atm)
V = Volume of O2 gas (in L)
n = Number of moles of O2 gas
R = Ideal gas constant (0.0821 L*atm/mol*K)
T = Temperature in Kelvin (25 C + 273.15 = 298.15 K)
n = (P * V) / (R * T)
n = (724.2 torr * 0.269 L) / (0.0821 L*atm/mol*K * 298.15 K)
Now, convert torr to atm:
n = (724.2 torr * 0.269 L) / (0.0821 L*atm/mol*K * 298.15 K) * (1 atm / 760 torr)
n = 0.0100 mol (approximately)
According to the balanced equation, 1 mole of KClO4 produces 2 moles of O2 gas. So, the moles of KClO4 decomposed would be half of the moles of O2 gas produced:
moles of KClO4 = 0.0100 mol / 2
moles of KClO4 = 0.00500 mol
Finally, to find the mass of potassium perchlorate that decomposed, we need to use the molar mass of KClO4. The molar mass of KClO4 is:
K = 39.10 g/mol
Cl = 35.45 g/mol
O = 16.00 g/mol (4 oxygen atoms)
Molar mass of KClO4 = (39.10 g/mol) + (35.45 g/mol) + (4 * 16.00 g/mol)
Molar mass of KClO4 = 138.55 g/mol
The mass of potassium perchlorate that decomposed is:
mass = moles * molar mass
mass = 0.00500 mol * 138.55 g/mol
Finally, calculate the mass:
mass = 0.6928 g
Therefore, approximately 0.6928 grams of potassium perchlorate decomposed.