If you have 0.56 moles of an ideal gas at 87 ° C and a pressure of 569 torr, what volume will the gas take up?
2 answers:
Explanation:
The given data is as follows.
n = 0.56 moles, T = = (87 + 273) K = 360 K
P = 569 torr = 0.748 atm (as 1 torr = 0.0013 atm)
According to ideal gas equation, PV = nRT
Therefore, putting given values into the above formula as follows.
PV = nRT
V = 22.12 L
Thus, we can conclude that the given gas will take up 22.12 L of volume.
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Answer:
118.22 atm
Explanation:
2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
KP = 0.13 =
Where p(SO₃) is the partial pressure of SO₃, p(SO₂) is the partial pressure of SO₂ and p(O₂) is the partial pressure of O₂.
- With 2.00 mol SO₂ and 2.00 mol O₂ if there was a 100% yield of SO₃, then 2 moles of SO₃ would be produced and 1.00 mol of O₂ would remain.
- With a 71.0% yield, there are only 2*0.71 = 1.42 mol SO₃, the moles of SO₂ that didn't react would be 2 - 1.42 = 0.58; and the moles of O₂ that didn't react would be 2 - 1.42/2 = 1.29.
The total number of moles is 1.42 + 0.58 + 1.29 = 3.29. With that value we can calculate the molar fraction (X) of each component:
- XSO₂ = 0.58/3.29 = 0.176
- XO₂ = 1.29/3.29 = 0.392
- XSO₃ = 1.42/3.29 = 0.432
The partial pressure of each gas is equal to the total pressure (PT) multiplied by the molar fraction of each component.
- p(SO₂) = 0.176 * PT
- p(O₂) = 0.392 * PT
- p(SO₃) = 0.432 * PT
Rewriting KP and solving for PT: