# Question 8c2bd

Aug 4, 2017

Here's what I got.

#### Explanation:

The balanced chemical equation that describes this reaction looks like this

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

Notice that the reaction consumes $2$ moles of carbon monoxide and produces $2$ moles of carbon dioxide for every $1$ mole of oxygen gas that takes part in the reaction.

Now, you know that all three gases are kept under the same conditions for pressure and temperature. This means that you can treat the mole ratios that exist between the three gases as volume ratios.

In other words, you can say that the reaction consumes ${\text{2 cm}}^{3}$ of carbon monoxide and produces ${\text{2 cm}}^{3}$ of carbon dioxide for every ${\text{1 cm}}^{3}$ of oxygen gas that takes part in the reaction.

So, in order for the reaction to consume all ${\text{50 cm}}^{3}$ of oxygen gas available, you would need

50 color(red)(cancel(color(black)("cm"^3color(white)(.)"O"_2))) * ("2 cm"^3color(white)(.)"CO")/(1color(red)(cancel(color(black)("cm"^3color(white)(.)"O"_2)))) = "100 cm"^3color(white)(.)"CO"

Since you don't have enough carbon monoxide to ensure that all the sample of oxygen gas can take part in the reaction

overbrace("100 cm"^3color(white)(.)"CO")^(color(blue)("what you need")) " " > " " overbrace("80 cm"^3color(white)(.)"CO")^(color(blue)("what you have"))

you can say that carbon monoxide will act as a limiting reagent, i.e. it will be completely consumed before all the sample of oxygen gas will get the chance to react.

Therefore, you can say that the reaction will consume ${\text{80 cm}}^{3}$ of carbon monoxide and

80 color(red)(cancel(color(black)("cm"^3color(white)(.)"CO"))) * ("1 cm"^3color(white)(.)"O"_2)/(2color(red)(cancel(color(black)("cm"^3color(white)(.)"CO")))) = "40 cm"^3color(white)(.)"O"_2

and produce

80 color(red)(cancel(color(black)("cm"^3color(white)(.)"CO"))) * ("2 cm"^3color(white)(.)"CO"_2)/(2color(red)(cancel(color(black)("cm"^3color(white)(.)"CO")))) = "80 cm"^3color(white)(.)"CO"_2#

Once the reaction is complete, you will be left with

${\text{50 cm"^3color(white)(.)"O"_2 - "40 cm"^3color(white)(.)"O"_2 = "10 cm"^3color(white)(.)"O}}_{2}$

The reaction vessel will contain

$\left\{\begin{matrix}\text{80 cm"^3color(white)(.)"CO"_2 color(white)(a)-> "produced by the reaction" \\ "10 cm"^3color(white)(.)"O"_2 color(white)(aa)-> "unreacted}\end{matrix}\right.$