#"1 L"# of #"CO"_2# is passed through red hot coke. The volume becomes #"1.4 L"# at the same temperature and pressure. What is the composition of the products?

1 Answer
Aug 19, 2017

Answer:

Here's what I got.

Explanation:

Start by writing the balanced chemical equation that describes this reaction

#"CO"_ (2(g)) + "C"_ ((s)) -> 2"CO"_ ((g))#

Now, notice that every mole of carbon dioxide that takes part in the reaction produces #2# moles of carbon monoxide.

This implies that if the sample of carbon dioxide would react completely, you would end up with #"2 L"# of carbon monoxide--keep in mind that when working with gases kept under the same conditions for pressure and temperature, their mole ratio is equivalent to a volume ratio.

In your case, the resulting gaseous mixture has a volume of #"1.4 L"#, so you know for a fact that the carbon dioxide was not completely consumed.

If you take #x# #"L"# to be the volume of carbon dioxide consumed by the reaction, you can say that #2x# #"L"# will be the volume of carbon dioxide produced by the reaction.

This means that after the reaction is complete, you will be left with

#overbrace((1 - x)color(white)(.)color(red)(cancel(color(black)("L"))))^(color(blue)("volume of unreacted CO"_2)) + overbrace((2x)color(white)(.)color(red)(cancel(color(black)("L"))))^(color(blue)("volume of CO produced")) = 1.4color(red)(cancel(color(black)("L")))#

This will get you

# 1 + x =1.4 implies x= 0.4#

Therefore, you can say that the resulting mixture will contain

#(1 - 0.4)color(white)(.)"L" = "0.6 L " -> " CO"_2#

#(2 * 0.4)color(white)(.)"L" = "0.8 L " -> " CO"#