# Question #ff991

##### 1 Answer

Here's what I got.

#### Explanation:

You can't calculate the *volume* of carbon dioxide that can be removed by this reaction because you don't have any information about the **pressure** and the **temperature** of the gas.

So at best, you can use the information you have to find the number of *moles* of carbon dioxide that can be removed by your reaction.

You know that the balanced chemical equation looks like this

#2"LiOH"_ ((aq)) + "CO"_ (2(g)) -> "Li"_ 2"CO"_ (3(s)) + "H"_ 2"O"_ ((l))#

This tells you that in order to absorb **mole** of carbon dioxide, you need to have **moles** of lithium hydroxide.

Use the molarity and the volume of the lithium hydroxide solution to find the number of moles of solute it contains

#3.45 color(red)(cancel(color(black)("L solution"))) * overbrace("0.10 moles LiOH"/(1color(red)(cancel(color(black)("L solution")))))^(color(blue)("= 0.10 M LIOH solution")) = "0.345 moles LiOH"#

You can use the **mole ratio** that exists between the two reactants to find the number of moles of carbon dioxide.

#0.345 color(red)(cancel(color(black)("moles LiOH"))) * overbrace("1 mole CO"_2/(2color(red)(cancel(color(black)("moles LiOH")))))^(color(blue)("given by the balanced chemical equation")) = "0.1725 moles CO"_2#

At this point, you would need to use the **ideal gas law equation**

#color(blue)(ul(color(black)(PV = nRT)))#

Here

#P# is the pressure of the gas#V# is the volume it occupies#n# is the number of moles of gas present in the sample#R# is theuniversal gas constant, equal to#0.0821("atm L")/("mol K")# #T# is theabsolute temperatureof the gas

to find the volume of carbon dioxide that would contain this many moles under the conditions you have for pressure and temperature.

#PV = nRT implies V = (nRT)/P#

If you have a pressure of

#V = (0.1725 color(red)(cancel(color(black)("moles"))) * 0.0821(color(red)(cancel(color(black)("atm"))) * "L")/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * T color(red)(cancel(color(black)("K"))))/(Pcolor(red)(cancel(color(black)("atm"))))#

#V = 0.01416 * T/P#