Question

What is the mass of `"5.6 L"` of `"O"_2"` at `23^@"C"` and `"8.16 atm"`?

Answer 1

The mass of `"O"_2` is `"60.2 g"`

Explanation:

This can be solved by using the Ideal Gas Law:

`PV=nRT`,

where `P` is pressure, `V` is volume, `n` is moles, `R` is the universal gas constant, and `T`, which is the temperature in Kelvins.

First calculate the number of moles of `"O"_2"`, using the equation for the Ideal Gas Law. Then multiply the moles `"O"_2` by its molar mass to get the mass of `"O"_2"`.

`color(blue)("Organize your data"`

Known/Given

`P="8.16 atm"`

`V="5.60 L"`

`R="0.08206 L atm K"^(-1) "mol"^(-1)`

`T="23"^@"C"+273="296 K"`

`color(blue)("Moles of Oxygen Gas"`

Rearrange the equation to isolate `n`. Insert your data and solve.

`n=(PV)/(RT)`

`n=(8.16color(red)cancel(color(black)("atm"))xx5.60color(red)cancel(color(black)("L")))/(0.08206 color(red)cancel(color(black)("L")) color(red)cancel(color(black)("atm")) color(red)cancel(color(black)("K"))^(-1) "mol"^(-1)xx296color(red)cancel(color(black)("K")))="1.88 mol"`

There are `"1.88 mol O"_2"` under the conditions described in the question.

`color(blue)("Mass of Oxygen Gas"`

Multiply the mol `"O"_2"` by its molar mass, `"31.998 g/mol"`.

`1.88color(red)cancel(color(black)("mol O"_2))xx(31.998"g O"_2)/(1color(red)cancel(color(black)("mol O"_2)))="60.2 g O"_2` rounded to three sig figs