Respuesta :
The inidividual partial pressure of all gases present afer reaction i.e. Xe₂, HF & O₂ is 1.52atm, 6.1 atm and 1.52 atm respectively.
How do we calculate partial pressure?
Partial pressure of each gas involved in the chemical reaction will be calculated by using the ideal gas equation with each moles seprately, as:
PV = nRT, where
P = partial pressure = ?
V = volume = 2L
n = moles of gases
R = universal gas constant = 0.082 L.atm / K.mol
T = temperature = 25°C = 298.15 K
Given balanced chemical reaction is:
2XeF₂(g) + 2H₂O(l) → Xe₂(g) + 4HF(g) + O₂(g)
Moles from mass will be calculated as:
n = W/M, where
W = given mass
M = molar mass
Moles of XeF₂ = 67.72g / 169.3g/mol = 0.4 mol
Moles of H₂O = 4.50g / 18g/mol = 0.25 mol
From the calculation it is clear that water is the limiting reagent so the formation of products will depends on water. From the stoichiometry of the reaction it is clear that:
0.25 moles of H₂O = produces 0.25/2 = 0.125 moles of Xe₂
0.25 moles of H₂O = produces 0.25(4/2) = 0.5 moles of HF
0.25 moles of H₂O = produces 0.25/2 = 0.125 moles of O₂
So partial pressure of oxygen and xenon gas will be same as they have same number of moles and it will be calculated as:
P = (0.125)(0.082)(298.15) / (2) = 1.52 atm
Partial pressure of HF gas will be calculated as:
P = (0.5)(0.082)(298.15) / (2) = 6.1 atm
Hence partial pressure of Xe₂, HF & O₂ is 1.52atm, 6.1 atm and 1.52 atm respectively.
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