# How many liters would 14.0 grams of chlorine gas occupy at 300.0 K and 1.51 atm?

Nov 7, 2015

$\text{3.22 L}$

#### Explanation:

The first thing to do here is use the molar mass of chlorine gas to determine how many moles of chlorine gas you have in that sample.

14.0color(red)(cancel(color(black)("g"))) * "1 mole Cl"_2/(70.906color(red)(cancel(color(black)("g")))) = "0.1974 moles Cl"_2

Now that you know how many moles of gas you're dealing with, and the conditions for pressure andtemperature at which the gas is kept, you can use the ideal gas law equation to solve for the volume of the sample.

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

$P$ - the pressure of the sample
$V$ - the volume it occupies
$n$ - the number of moles of gas
$R$ - the universal gas constant, usually given as $0.082 \left(\text{atm" * "L")/("mol" * "K}\right)$
$T$ - the temperature of the gas, expressed in Kelvin

Rearrange this equation to solve for $V$, then plug in your values to get

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

V = (0.1974color(red)(cancel(color(black)("moles"))) * 0.082(color(red)(cancel(color(black)("atm"))) * "L")/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * 300.0color(red)(cancel(color(black)("K"))))/(1.51color(red)(cancel(color(black)("atm")))) = "3.2159 L"

Rounded to three sig figs, the answer will be

$V = \textcolor{g r e e n}{\text{3.22 L}}$