Final answer:
The mass of the oxygen gas obtained from the decomposition of KClO3, collected at 24°C and a pressure of 783 mm Hg is approximately 0.234 grams. This calculation takes into account the subtraction of water vapor pressure from atmospheric pressure and uses the Ideal Gas Law.
Step-by-step explanation:
The mass of oxygen gas obtained in decomposition of KClO3 can be calculated using the Ideal Gas Law equation, which is PV = nRT. First, we must account for the partial pressure of the oxygen gas by subtracting the water vapor pressure from the atmospheric pressure. Then, we use the Ideal Gas Law to find the number of moles of oxygen gas and convert those moles to mass.
Since we have a mix of gases, we subtract the water vapor pressure from the total pressure to get the pressure due to oxygen gas alone:
Oxygen gas pressure = 783 mm Hg - 22.4 mm Hg = 760.6 mm Hg
Now convert this pressure to atmospheres (1 atm = 760 mm Hg):
Oxygen gas pressure = 760.6 mm Hg / 760 mm Hg/atm ≈ 1 atm
To find the number of moles (n) of oxygen using the Ideal Gas Law, we must use R (the universal gas constant), which is 0.0821 L atm/mol K, and convert the temperature to Kelvin (K).
T(K) = 24°C + 273 = 297 K
Thus:
PV = nRT
1 atm × 0.179 L = n × 0.0821 L atm/mol K × 297 K
Solving for n:
n = (1 atm × 0.179 L) / (0.0821 L atm/mol K × 297 K)
n ≈ 0.0073 moles
To find the mass of oxygen, we multiply the number of moles by the molar mass of O2, which is 32 g/mol:
Mass = moles × molar mass
Mass = 0.0073 moles × 32 g/mol
Mass ≈ 0.234 g
Therefore, the mass of oxygen gas obtained is approximately 0.234 grams.