First, let's calculate the partial pressure of water at 22 degrees Celsius using the vapor pressure of water at that temperature:
Partial pressure of water = total pressure - vapor pressure of water
Partial pressure of water = 755 toor - 21 toor
Partial pressure of water = 734 toor
Next, let's calculate the partial pressure of oxygen in the gas collected:
Partial pressure of oxygen = total pressure - partial pressure of water
Partial pressure of oxygen = 755 toor - 734 toor
Partial pressure of oxygen = 21 toor
Now, let's convert the volume of oxygen collected to moles using the ideal gas law:
PV = nRT
n = PV/RT
n = (21 toor * 0.650L) / (0.0821 L.atm/mol.K * 295 K)
n = 0.56 mol
Since the reaction is 2KClO3 -> 2KCl + 3O2, we can see that 2 moles of KClO3 produce 3 moles of O2. Therefore, 0.56 moles of O2 would have required (0.56 * 2/3) moles of KClO3 to decompose:
0.56 * 2/3 = 0.373 moles of KClO3
Finally, let's calculate the mass of KClO3 in the sample that was decomposed:
Molar mass of KClO3 = 39.10 g/mol + 35.45 g/mol + 16.00 g/mol*3 = 122.55 g/mol
Mass of KClO3 = 0.373 moles * 122.55 g/mol = 45.72 g
Therefore, the partial pressure of oxygen in the gas collected was 21 toor and the mass of KClO3 in the sample that was decomposed was 45.72 g.
Based on the reaction below
2KClO3(s) = 2KCl(s) + 3O(g)
The oxygen produce was collected by the displacement of water at 22 degrees Celsius at total pressure of 755 toor. Volume of gas collected was 0.650L. And the vapor pressure of water at 22 degrees Celsius is 21toor .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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