Answer:
The composition of the original mixture in molepercent is 80% of H₂ and 20% of O₂.
Step-by-step 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.