How many liters of ozone can be destroyed at 220. K and 5.0 kPa if 200.0 g of chlorine gas react with it?

Chlorine in the upper atmosphere can destroy ozone molecules, ${O}_{3}$. The reaction can be represented by the following equation: $C {l}_{2} \left(g\right) + 2 {O}_{3} \left(g\right) \to 2 C l O \left(g\right) + 2 {O}_{2} \left(g\right)$

Jun 27, 2018

$\text{200.0 g Cl"_2}$ can destroy $\text{2100 L O"_3}$.

Explanation:

Balanced equation

$\text{Cl"_2("g") + "2O"_3("g")}$$\rightarrow$$\text{2ClO(g) + 2O"_2("g")}$

First we need to determine how many moles of ozone will react with $\text{200.0 g Cl"_2}$. Then we'll use the equation for the ideal gas law to determine the volume of ozone that will be destroyed.

Determine mol $\text{Cl"_2}$ by dividing its given mass by its molar mass $\left(\text{68.9 g/mol}\right)$. Do this by multiplying by the inverse of the molar mass (mol/g). Then determine mol $\text{O"_3}$ by multiplying by the mol ratio between ozone and chlorine gas in the balanced equation, with ozone in the numerator.

200.0color(red)cancel(color(black)("g Cl"_2))xx(1color(red)cancel(color(black)("mol Cl"_2)))/(68.9color(red)cancel(color(black)("g Cl"_2)))xx("mol O"_3)/(1color(red)cancel(color(black)("mol Cl"_2)))="5.81 mol O"_3"

Ideal gas law equation

$P V = n R T$

Known

$P = \text{5.0 kPa}$

$n = \text{5.81 mol O"_3}$

$R = \text{8.31447 L kPa K"^(-1) "mol"^(-1)}$
https://www.katmarsoftware.com/gconvals.htm

$T = \text{220. K}$

Unknown

$V$

Solution

Rearrange the equation to isolate volume, $V$. Plug in the known values and solve.

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

V=(5.81color(red)cancel(color(black)("mol"))xx8.31447" L" color(red)cancel(color(black)("kPa"))color(red)cancel(color(black)( "K"))^(-1) color(red)cancel(color(black)("mol"))^(-1)xx220color(red)cancel(color(black)("K")))/(5.0color(red)cancel(color(black)("kPa")))="2100 L"
(rounded to two significant figures due to $\text{5.0 kPa}$)

$\text{200.0 g Cl"_2}$ can destroy $\text{2100 L O"_3}$.