Answer:
The composition of the original mixture in molepercent is 80% of Hâ‚‚ and 20% of Oâ‚‚.
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 (Hâ‚‚) + P (Oâ‚‚)
The number of moles can be found by Pt = nt RT/V, in which nt = n (Hâ‚‚) +n (Oâ‚‚).
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 Oâ‚‚ we need two mol of Hâ‚‚.
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 Hâ‚‚ remains in the system.
This means that in the beginning we have n mol of Hâ‚‚, and when x mol of Hâ‚‚ reacts with 0,5x mol of Oâ‚‚, 0.4 mol of Hâ‚‚ 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 Hâ‚‚ reacted with 0.2 mol of Oâ‚‚ and 0.4 mol of Hâ‚‚ remained as excess.
Therefore, in the beginning we had 0.8 mol of Hâ‚‚ and 0.2 mol of Oâ‚‚. Thus the molepercent of the mixture is 80% of Hâ‚‚ and 20% of Oâ‚‚.
