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

Jun 1, 2017

The mass of ${\text{O}}_{2}$ is $\text{60.2 g}$

#### Explanation:

This can be solved by using the Ideal Gas Law:

$P V = n R T$,

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 $\text{O"_2}$, using the equation for the Ideal Gas Law. Then multiply the moles ${\text{O}}_{2}$ by its molar mass to get the mass of $\text{O"_2}$.

color(blue)("Organize your data"

Known/Given

$P = \text{8.16 atm}$

$V = \text{5.60 L}$

$R = {\text{0.08206 L atm K"^(-1) "mol}}^{- 1}$

$T = \text{23"^@"C"+273="296 K}$

color(blue)("Moles of Oxygen Gas"

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

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

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 $\text{1.88 mol O"_2}$ under the conditions described in the question.

color(blue)("Mass of Oxygen Gas"

Multiply the mol $\text{O"_2}$ by its molar mass, $\text{31.998 g/mol}$.

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