# What is the volume of 5.0 grams of CO_2 at STP?

May 18, 2017

The volume of $\text{5.0 g CO"_2}$ is $\text{2.6 L CO"_2}$ at STP.

#### Explanation:

color(blue)("STP"

STP is currently ${0}^{\circ} \text{C}$ or $\text{273.15 K}$, which are equal, though the Kelvin temperature scale is used for gas laws; and pressure is $\text{10"^5color(white)(.)"Pascals (Pa)}$, but most people use $\text{100 kPa}$, which is equal to ${10}^{5} \textcolor{w h i t e}{.} \text{Pa}$.

You will use the ideal gas law to answer this question. Its formula is:

$P V = n R T$,

where $P$ is pressure, $V$ is volume, $n$ is moles, $R$ is a gas constant, and $T$ is temperature in Kelvins.

color(blue)("Determine moles"
You may have noticed that the equation requires moles $\left(n\right)$, but you have been given the mass of $\text{CO"_2}$. To determine moles, you multiply the given mass by the inverse of the molar mass of $\text{CO"_2}$, which is $\text{44.009 g/mol}$.

5.0color(red)cancel(color(black)("g CO"_2))xx(1"mol CO"_2)/(44.009color(red)cancel(color(black)("g CO"_2)))="0.1136 mol CO"_2"

color(blue)("Organize your data".

Given/Known
$P = \text{100 kPa}$
$n = \text{0.1136 mol}$
$R = \text{8.3145 L kPa K"^(-1) "mol"^(-1)}$
https://en.wikipedia.org/wiki/Gas_constant
$T = \text{273.15 K}$

Unknown: $V$

color(blue)("Solve for volume using the ideal gas law."
Rearrange the formula to isolate $V$. Insert your data into the equation and solve.

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

V=(0.1136color(red)cancel(color(black)("mol"))xx8.3145 color(white)(.)"L" color(red)cancel(color(black)("kPa")) color(red)cancel(color(black)("K"))^(-1) color(red)cancel(color(black)("mol"))^(-1)xx273.15color(red)cancel(color(black)("K")))/(100color(red)cancel(color(black)("kPa")))="2.6 L CO"_2" rounded to two significant figures due to $\text{5.0 g}$

May 18, 2017

I got 2.55 Liters

#### Explanation:

1 mole of any gas at STP = 22.4 Liters
5 g CO_2(g) = (5 g)/(44(g/"mole")) = 0.114 "mole" CO_2(g)
Volume of 0.114 mole $C {O}_{2} \left(g\right)$ = (0.114 mole)(22.4 L/mole) = 2.55 Liters $C {O}_{2}$(g) at STP