# What is the mole fraction of NO in a 55.0 L gas cylinder at 30.0°C which comes from a mixture of N_2 and NO if you have 3.238 mol of N_2 and the gas cylinder has a total pressure of 2.14 atm?

Nov 27, 2015

$0.315$

#### Explanation:

Your strategy here will be to use the ideal gas law equation in order to find the total number of moles present in the mixture, then use the number of moles of nitrogen gas to determine the mole fraction of nitric oxide.

So, the ideal gas law equation looks like this

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

$P$ - the pressure of the gas
$V$ - the volume it occupies
$n$ - the number of moles of gas
$R$ - the universal gas constant, usually given as $0.0821 \left(\text{atm" * "L")/("mol" * "K}\right)$
$T$ - the temperature of the gas, expressed in Kelvin

Plug in your values and solve for $n$ to get

$P V = n R T \implies n = \frac{P V}{R T}$

n = (2.14 color(red)(cancel(color(black)("atm"))) * 55.0color(red)(cancel(color(black)("L"))))/(0.0821 (color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 30.0)color(red)(cancel(color(black)("K")))) = "4.729 moles"

Since the mixture only contains nitric oxide and nitrogen gas, it follows that the number of moles of nitric acid will be equal to

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

${n}_{N O} = 4.729 - 3.238 = \text{1.491 moles NO}$

Now, the mole fraction of nitric oxide will be equal to the number of moles of nitric oxide divided by the total number of moles present in the mixture.

${\chi}_{\text{NO" = n_"NO"/n_"total}}$

In this case, you have

chi_"NO" = (1.491 color(red)(cancel(color(black)("moles"))))/(4.729 color(red)(cancel(color(black)("moles")))) = color(green)(0.315)

The answer is rounded to three sig figs.