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)#

1 Answer
Jun 27, 2018

Answer:

#"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"#.