First, we need to calculate the partial pressure of oxygen in the gas collected.
To do this, we need to consider Dalton's law of partial pressures, which states that the total pressure of a mixture of gases is equal to the sum of the partial pressures of the individual gases.
Since the total pressure is 754 torr and the vapor pressure of water is 21 torr, the partial pressure of oxygen can be calculated as follows:
Partial pressure of oxygen = Total pressure - Vapor pressure of water
Partial pressure of oxygen = 754 torr - 21 torr
Partial pressure of oxygen = 733 torr
Next, we need to calculate the moles of oxygen gas collected using the ideal gas law equation:
PV = nRT
where
P = partial pressure of oxygen gas (733 torr)
V = volume of gas collected (0.650 L)
n = number of moles of gas
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature in Kelvin (22°C = 295 K)
Rearranging the ideal gas law equation to solve for n, we have:
n = PV / RT
n = (733 torr * 0.650 L) / (0.0821 L.atm/mol.K * 295 K)
n ≈ 18.3 mol
Since the reaction is balanced in a 1:3 ratio of KClO3 to O2, the number of moles of KClO3 decomposed is 1/2 of the number of moles of oxygen gas produced. Therefore, the number of moles of KClO3 decomposed is approximately 9.15 mol.
Finally, we can calculate the mass of KClO3 decomposed using the molar mass of KClO3 (122.55 g/mol):
Mass = number of moles * molar mass
Mass = 9.15 mol * 122.55 g/mol
Mass ≈ 1122 grams
Therefore, the partial pressure of oxygen in the gas collected is approximately 733 torr, and the mass of KClO3 in the sample that was decomposed is approximately 1122 grams.
In reaction of
2KClO3(s) = 2KCl(s) +3O2(g)
The oxygen produced was collected by the displacement of water at 22 degrees Celsius at a total pressure of 754 toor. The volume of the gas Collected was 0.650L, and the vapor pressure of water at 22 degrees Celsius is 21 toor. Calculate the partial pressure of oxygen in the gas collected and the mass of KClO3 in the sample that was decomposed.
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