Question

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: `Cl_2(g) + 2O_3(g) -> 2ClO(g) +2O_2(g)`

Answer 1

`"200.0 g Cl"_2"` can destroy `"2100 L O"_3"`.

Explanation:

Balanced equation

`"Cl"_2("g") + "2O"_3("g")"` `rarr` `"2ClO(g) + 2O"_2("g")"`

First we need to determine how many moles of ozone will react with `"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 `"Cl"_2"` by dividing its given mass by its molar mass `("68.9 g/mol")`. Do this by multiplying by the inverse of the molar mass (mol/g). Then determine mol `"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

`PV=nRT`

Known

`P="5.0 kPa"`

`n="5.81 mol O"_3"`

`R="8.31447 L kPa K"^(-1) "mol"^(-1)"`
https://www.katmarsoftware.com/gconvals.htm

`T="220. K"`

Unknown

`V`

Solution

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

`V=(nRT)/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 `"5.0 kPa"`)

`"200.0 g Cl"_2"` can destroy `"2100 L O"_3"`.