# What is the the volume occupied by 2.34 g of carbon dioxide gas at STP?

May 6, 2016

$V = 1.18 L$

#### Explanation:

In order to solve this problem we would use the Ideal Gas Law formula $P V = n R T$

$P =$ Pressure in $a t m$
$V =$ Volume in $L$
$n =$ moles
$R =$ Ideal Gas Law Constant
$T =$ Temp in $K$

$S T P$ is Standard Temperature and Pressure which has values of
$1 a t m$ and $273 K$

$2.34 g C {O}_{2}$ must be converted to moles

$2.34 g C {O}_{2} x \frac{1 m o l}{44 g C {O}_{2}} = 0.053 m o l s$

$P = 1 a t m$
V= ??? L
$n = 0.053 m o l s$
$R = 0.0821 \frac{a t m L}{m o l K}$
$T = 273 K$

$P V = n R T$ becomes

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

$V = \frac{0.053 \cancel{m o l s} \left(0.0821 \frac{\cancel{a t m} L}{\cancel{m o l} \cancel{K}}\right) \left(273 \cancel{K}\right)}{1 a t m}$

$V = 1.18 L$

May 7, 2016

$1.18 L$

#### Explanation:

Alternative solution:

Calculate how many moles $2.34 g$ $C {O}_{2}$ is

$2.34 g \cdot \frac{1 m o l}{44 g C {O}_{2}} = 0.053 m o l$

Simple mole ratios tell us that $1$ mol of any gas at STP is $22.4 L$, but since we only have $0.053 m o l$, we times $22.4 \frac{L}{m o l}$ by $0.053 m o l .$

$\frac{22.4 L}{\cancel{m o l}} \cdot 0.053 \cancel{m o l} = 1.18 L$