# 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?

Feb 6, 2016

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

${n}_{\text{total}} = {n}_{{O}_{2}} + {n}_{{N}_{2}} + {n}_{{H}_{2}}$

${n}_{\text{total" = 0.47127 + 0.29165 + 1.3096 = "2.0725 moles}}$

So, to get the mole fraction of a component $i$ of the mixture, all you have to do is divide the number of moles of that component by the total number of moles present.

$\textcolor{b l u e}{{\chi}_{i} = {n}_{i} / {n}_{\text{total}}}$

You will have

${\chi}_{{O}_{2}} = \left(0.47127 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles"))))/(2.0725color(red)(cancel(color(black)("moles}}}}\right) = \textcolor{g r e e n}{0.227}$

${\chi}_{{N}_{2}} = \left(0.29165 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles"))))/(2.0725color(red)(cancel(color(black)("moles}}}}\right) = \textcolor{g r e e n}{0.141}$

${\chi}_{{H}_{2}} = \left(1.3096 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles"))))/(2.0725color(red)(cancel(color(black)("moles}}}}\right) = \textcolor{g r e e n}{0.632}$

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

$\textcolor{b l u e}{P V = n R T}$

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"))))

${P}_{\text{total" = "3.163 atm}}$

The partial pressure of each gas $i$ will depend on its mole fraction and on the total pressure of the gas

$\textcolor{b l u e}{{P}_{i} = {\chi}_{i} \times {P}_{\text{total}}}$

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.