Question

When a container is filled with 3.00 moles of H2, 2.00 moles of O2, and 1.00 mole of N2, the pressure in the container is 768 kPa. What is the partial pressure of O2? 154 kPa 192 kPa 128 kPa 256 kPa

Answer 1

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 2

Answer:

256kPa

Explanation:

You can find the partial pressure applying the Dalton´s law of partial pressure, that is:

$P{i}=n_{i}P_{T}$

where $P_{i}$ is the partial pressure of each gas, $n_{i}$ is the molar fraction of the gas and $P_{T}$ is the total pressure.

To found these values, first of all you should find the molar fraction of each gas:

For $H_{2}$:

$n_{H_{2}}=\frac{3.00molesofH_{2}}{3.00molesH_{2}+2.00molesO_{2}+1.00molN_{2}}$

$n_{H_{2}}=0.5$moles of $H_{2}$

For $O_{2}$:

$n_{O_{2}}=\frac{2.00molesofO_{2}}{3.00molesH_{2}+2.00molesO_{2}+1.00molN_{2}}$

$n_{O_{2}}=0.33333$moles of $O_{2}$

For $N_{2}$:

$n_{N_{2}}=\frac{1.00molesofN_{2}}{3.00molesH_{2}+2.00molesO_{2}+1.00molN_{2}}$

$n_{N_{2}}=0.166667$moles of $N_{2}$

Then you can apply the Dalton´s law, replacing the values for the molar fraction :

For $H_{2}$:

$P_{H_{2}}=n_{H_{2}}*P_{T}$

$P_{H_{2}}=0.5mol*768kPa$

$P_{H_{2}}=384kPa$

For $O_{2}$:

$P_{O_{2}}=n_{O_{2}}*P_{T}$

$P_{O_{2}}=0.33333mol*768kPa$

$P_{O_{2}}=256kPa$

For $N_{2}$:

$P_{N_{2}}=n_{N_{2}}*P_{T}$

$P_{N_{2}}=0.166667mol*768kPa$

$P_{N_{2}}=128kPa$