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A mass of 7.75 g carbon monoxide is reacted with 8.04 g hydrogen to form methanol. CO(g) + 2 H2(g) → CH3OH(g) The reaction is performed in a 5.00 L flask at 85.0 °C and proceeds to completion. What is the partial pressure (in mmHg) of each of the three species following completion of the reaction? What is the total pressure in the flask? CO =? mmHg H2 =? mmHg CH3OH =? mmHg Total =? mmHg

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Answer:

Pco(g) = 47934.93 mmHg

PH2(g) = 405538.12 mmHg

PCH3OH(g) = 105.07 mmHg

Ptotal = 453577.62 mmHg

Explanation:

CO(g) + 2H2(g) ↔ CH3OH (g)

7.75g     8.04g

V = 5L;  T = 85°C

⇒ Raoult's law:

  • Pa = Xa * P°a

∴ Pa : partial pressure of component a

∴ Xa : Molar fraction of component a in the mixture

∴ P°a : vapor pressure of the pure component a.

from literature:

  • Mw CO(g) = 28.0 g/mol
  • Mw H2(g) = 2.016 g/mol
  • Mw CH3OH(g) = 32.04 g/mol
  • P°co (85°C) = 737460.54 mmHg ......using Antoine constant 
  • P°H2  (85°C) = 433730.607 mmHg
  • P°CH3OH(g) (85°C) = 1616.533 mmHg

∴  Xco = ( mol CO(g) ) / total mol

⇒ mol CO(g) = 7.75g * ( mol / 28 g ) = 0.2768 mol CO(g)....limit reagent

⇒ mol H2(g) = 8.04g * ( mol / 2.016 g ) = 3.988 mol H2(g)

⇒ total mol = 0.2768 + 3.988 = 4.265 mol mixture

⇒ Xco(g) = 0.2768 / 4.265 = 0.065

⇒ XH2(g) = 3.988 / 4.265 = 0.935

⇒ Pco = 0.065 * 737460.54 mmHg = 47934.93 mmHg

⇒ PH2(g) = 0.935 * 433730.607 mmHg = 405538.12 mmHg

∴ mol CH3OH(g) = 0.2768 mol CO(g) * ( mol CH3OH / mol CO(g) ) = 0.2768 mol

⇒ X CH3OH(g) = 0.2768 / 4.265 = 0.065

⇒ PCH3OH(g) = 0.065 * 1616.533 mmHg = 105.07 mmHg

Dalton's law:

  • Pt = ∑ Pa

⇒ Pt = Pco(g) + PH2(g) + PCH3OH(g)

⇒ Pt = 47934.43 + 405538.12 + 105.07

⇒ Pt = 453577.62 mmHg

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