Question #ff991

1 Answer
Jul 12, 2017

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 #1# mole of carbon dioxide, you need to have #2# 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 #2:1# 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 the universal gas constant, equal to #0.0821("atm L")/("mol K")#
  • #T# is the absolute temperature of 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 #P# #"atm"# and a temperature of #T# #"K"#, then the volume of the gas would be equal to

#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#