Answer:
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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![[CH3CH2COO^-] ](https://tex.z-dn.net/?f=%5BCH3CH2COO%5E-%5D%0A)
= 0.05![[CH3CH2COOH]](https://tex.z-dn.net/?f=%5BCH3CH2COOH%5D)
= 0.10![\frac{[H^+][0.05]}{[0.1]}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5BH%5E%2B%5D%5B0.05%5D%7D%7B%5B0.1%5D%7D%20)
![[H^+] = \frac{(1.259*10^-5)(0.1)}{0.05}](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%20%3D%20%20%5Cfrac%7B%281.259%2A10%5E-5%29%280.1%29%7D%7B0.05%7D%20%20)
![[H^+]= 2.513*10^-5](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D%202.513%2A10%5E-5)
] = -log(2.513*10^-5)= 4.59.