# 10 cm^3 of carbon monoxide and 10 cm^3 of oxygen are mixed. What will be the total volume of gas present, in cm^3, after the reaction?

## They are mixed and the volume is measured at the same temperature and pressure. $2 C O + {O}_{2} \to 2 C {O}_{2}$ a. 10 b. 15 c. 20 d. 25 The answer given is b. 15, why?

Feb 16, 2016

The volume of gas will indeed be equal to ${\text{15 cm}}^{3}$.

#### Explanation:

This one is a little tricky. And here's why I say that.

When it comes to reaction that involve gases kept under the same conditions for pressure and temperature, it's important to realize that the mole ratios that exist between the species involved in the reaction becomes equivalent to the volume ratio.

So, take a look at the balanced chemical equation for your reaction

$\textcolor{red}{2} {\text{CO"_text((g]) + "O"_text(2(g]) -> color(blue)(2)"CO}}_{\textrm{2 \left(g\right]}}$

Notice that you have a $\textcolor{red}{2} : 1$ mole ratio between carbon monoxide and oxygen gas. This tells you that the reaction will always consume twice as many moles o carbon monoxide than of oxygen gas.

But since we're working with gases under the same conditions for pressure an temperature, you can say the exact same thing about their volumes.

In other words, the reaction will always consume a volume of carbon monoxide that is twice as large as the volume of oxygen gas.

Now, here's where the tricky part comes in. Notice that the problem provides you with equal volumes of $\text{CO}$ and ${\text{O}}_{2}$.

This is a problem because you know that you'd need a volume of $\text{CO}$ that is twice as large as the volume of ${\text{O}}_{2}$.

You an thus conclude that you're dealing with a limiting reagent. More specifically, carbon monoxide will act as a limiting reagent because it will be consumed before all the oxygen gets a chance to react.

So, you can say that the reaction will only consume

10color(red)(cancel(color(black)("cm"^3"CO"))) * ("1 cm"^3"O"_2)/(color(red)(2)color(red)(cancel(color(black)("cm"^3"CO")))) = "5 cm"^3color(white)(a)"O"_2

So, ${\text{10 cm}}^{3}$ of $\text{CO}$ wil lreact with ${\text{5 cm}}^{3}$ of ${\text{O}}_{2}$, leaving ${\text{5 cm}}^{3}$ of ${\text{O}}_{2}$ in excess.

Now look at the $\textcolor{red}{2} : \textcolor{b l u e}{2}$ volume ratio that exists between $\text{CO}$ and ${\text{CO}}_{2}$. This tells you that for a given volume of $\text{CO}$ that takes part in the reaction, the reaction produces an equal volume of ${\text{CO}}_{2}$.

So, if ${\text{10 cm}}^{3}$ of $\text{CO}$ react, you will get ${\text{10 cm}}^{3}$ of ${\text{CO}}_{2}$ produced.

This means that after the reaction is finished, the reaction vessel will contain

V_"gas" = overbrace(V_(O_2))^(color(purple)("in excess")) + overbrace(V_(CO_2))^(color(brown)("produced"))

V_"gas" = "5 cm"^3 + "1 0cm"^3 = color(green)("15 cm"^3)