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?

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.

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm