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4 February, 17:04

The total pressure of a mixture of oxygen and hydrogen

is1.00atm. The mixture is ignited and the water is removed.

Theremaining gas is pure hydrogen and excerts a pressure of. 4atm

whenmeasured at the same values of T and V as the original

mixture. What was the composition of the original mixture in

molepercent?

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Answers (1)
  1. 4 February, 17:43
    0
    The composition of the original mixture in molepercent is 80% of H₂ and 20% of O₂.

    Explanation:

    We need to combine the ideal gas law (PV = nRT) and Dalton's law of partial pressure (Pt = Pa + Pb + Pc + ...).

    The total pressure of the mixture is Pt = P (H₂) + P (O₂)

    The number of moles can be found by Pt = nt RT/V, in which nt = n (H₂) + n (O₂).

    If Pt is 1 atm, nt is 1.0 mol.

    Now we need to consider the chemical reaction below:

    H₂ + 0.5O₂ → H₂O

    This shows that for each mol of O₂ we need two mol of H₂.

    We know that the remaining gas is pure hydrogen and that its pressure is 0.4atm. Since PV = nRT, by the end of the reaction, 0.4 mol of H₂ remains in the system.

    This means that in the beginning we have n mol of H₂, and when x mol of H₂ reacts with 0,5x mol of O₂, 0.4 mol of H₂ reamains.

    If we have 1 mol in the begining and 0.4 mol in the end, the total amount of gas that reacted (x + 0.5X) is equal to 0.6 mol

    x + 0.5X = 0.6 mol ∴ x = 0.6 mol / 1.5 ∴ x = 0.4 mol

    0.4 mol of H₂ reacted with 0.2 mol of O₂ and 0.4 mol of H₂ remained as excess.

    Therefore, in the beginning we had 0.8 mol of H₂ and 0.2 mol of O₂. Thus the molepercent of the mixture is 80% of H₂ and 20% of O₂.
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