Respuesta :
The partial pressure of O2= 256 kpa
calculation
- The partial pressure of O2 =[ (moles of O2 / total moles ) x total pressure]
- total pressure = 768 kpa
- moles for O2= 2.00 moles
- total moles = Moles of O2 + moles of H2 + moles of N2
that is 3.00 moles + 2.00 moles + 1.00 moles = 6 .00 moles
- partial pressure= 2.00 moles/ 6.00 moles x 768 KPa = 256 KPa
Answer:
256kPa
Explanation:
You can find the partial pressure applying the Dalton´s law of partial pressure, that is:
[tex]P{i}=n_{i}P_{T}[/tex]
where [tex]P_{i}[/tex] is the partial pressure of each gas, [tex]n_{i}[/tex] is the molar fraction of the gas and [tex]P_{T}[/tex] is the total pressure.
To found these values, first of all you should find the molar fraction of each gas:
For [tex]H_{2}[/tex]:
[tex]n_{H_{2}}=\frac{3.00molesofH_{2}}{3.00molesH_{2}+2.00molesO_{2}+1.00molN_{2}}[/tex]
[tex]n_{H_{2}}=0.5[/tex]moles of [tex]H_{2}[/tex]
For [tex]O_{2}[/tex]:
[tex]n_{O_{2}}=\frac{2.00molesofO_{2}}{3.00molesH_{2}+2.00molesO_{2}+1.00molN_{2}}[/tex]
[tex]n_{O_{2}}=0.33333[/tex]moles of [tex]O_{2}[/tex]
For [tex]N_{2}[/tex]:
[tex]n_{N_{2}}=\frac{1.00molesofN_{2}}{3.00molesH_{2}+2.00molesO_{2}+1.00molN_{2}}[/tex]
[tex]n_{N_{2}}=0.166667[/tex]moles of [tex]N_{2}[/tex]
Then you can apply the Dalton´s law, replacing the values for the molar fraction :
For [tex]H_{2}[/tex]:
[tex]P_{H_{2}}=n_{H_{2}}*P_{T}[/tex]
[tex]P_{H_{2}}=0.5mol*768kPa[/tex]
[tex]P_{H_{2}}=384kPa[/tex]
For [tex]O_{2}[/tex]:
[tex]P_{O_{2}}=n_{O_{2}}*P_{T}[/tex]
[tex]P_{O_{2}}=0.33333mol*768kPa[/tex]
[tex]P_{O_{2}}=256kPa[/tex]
For [tex]N_{2}[/tex]:
[tex]P_{N_{2}}=n_{N_{2}}*P_{T}[/tex]
[tex]P_{N_{2}}=0.166667mol*768kPa[/tex]
[tex]P_{N_{2}}=128kPa[/tex]
