# "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?

Aug 19, 2017

Here's what I got.

#### Explanation:

Start by writing the balanced chemical equation that describes this reaction

${\text{CO"_ (2(g)) + "C"_ ((s)) -> 2"CO}}_{\left(g\right)}$

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 $\text{2 L}$ 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 $\text{1.4 L}$, so you know for a fact that the carbon dioxide was not completely consumed.

If you take $x$ $\text{L}$ to be the volume of carbon dioxide consumed by the reaction, you can say that $2 x$ $\text{L}$ 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

$\left(1 - 0.4\right) \textcolor{w h i t e}{.} {\text{L" = "0.6 L " -> " CO}}_{2}$

$\left(2 \cdot 0.4\right) \textcolor{w h i t e}{.} \text{L" = "0.8 L " -> " CO}$