# What are the mole fractions of each component in a mixture of 15.08 g of #O_2#, 8.17 g of #N_2#, and 2.64 g of #H_2#? What is the partial pressure in atm of each component of this mix ture if it is held in a 15.50-L vessel at 15°C?

##### 1 Answer

#### Answer:

Here's what I got.

#### Explanation:

In order to be able to calculate the **mole fraction** of each component in the mixture, you will have to convert the masses given to you to *moles*.

To do that, use the **molar mass** of oxygen gas, nitrogen gas, and hydrogen gas, respectively

#15.08 color(red)(cancel(color(black)("g"))) * "1 mole O"_2/(31.9988color(red)(cancel(color(black)("g")))) = "0.47127 moles O"_2#

#8.17 color(red)(cancel(color(black)("g"))) * "1 mole N"_2/(28.0134color(red)(cancel(color(black)("g")))) = "0.29165 moles N"_2#

#2.64color(red)(cancel(color(black)("g"))) * "1 mole H"_2/(2.0159color(red)(cancel(color(black)("g")))) = "1.3096 moles H"_2#

The **total number of moles** present in the mixture will be

#n_"total" = n_(O_2) + n_(N_2) + n_(H_2)#

#n_"total" = 0.47127 + 0.29165 + 1.3096 = "2.0725 moles"#

So, to get the mole fraction of a component

#color(blue)(chi_i = n_i/n_"total")#

You will have

#chi_(O_2) = (0.47127 color(red)(cancel(color(black)("moles"))))/(2.0725color(red)(cancel(color(black)("moles")))) = color(green)(0.227)#

#chi_(N_2) = (0.29165color(red)(cancel(color(black)("moles"))))/(2.0725color(red)(cancel(color(black)("moles")))) = color(green)(0.141)#

#chi_(H_2) = (1.3096color(red)(cancel(color(black)("moles"))))/(2.0725color(red)(cancel(color(black)("moles")))) = color(green)(0.632)#

Now, you *could* calculate the partial pressure of each gas by using the **ideal gas law** equation *three times*, once for every gas.

#color(blue)(PV = nRT)#

However, a simpler approach would be to use the ideal gas law equation *once* to find the **total pressure** exerted by the mixture, then use the mole fractions of the gases to find their respective partial pressure - think **Raoult's Law**.

So, the pressure exerted by the mixture will be

#P_"total" = (n_"total" * RT)/V#

#P_"total" = (2.0725 color(red)(cancel(color(black)("moles"))) * 0.0821("atm" * color(red)(cancel(color(black)("L"))))/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * (273.15 + 15)color(red)(cancel(color(black)("K"))))/(15.50color(red)(cancel(color(black)("L"))))#

#P_"total" = "3.163 atm"#

The partial pressure of each gas

#color(blue)(P_i = chi_i xx P_"total")#

Plug in your values to get

#P_(O_2) = 0.227 * "3.163 atm" = color(green)("0.718 atm")#

#P_(N_2) = 0.141 * "3.163 atm" = color(green)("0.446 atm")#

#P_(H_2) = 0.632 * "3.163 atm" ~~ color(green)("2.00 atm")#

All the answers are rounded to three sig figs.