# What are the volumes of each gas in the resulting mixture after the combustion of "20 cm"^3 color(white)(l)"CO" with "30 cm"^3color(white)(l)"O"_2?

Apr 13, 2017

The resulting gas consists of ${\text{20 cm"^3 color(white)(l) "CO}}_{2}$ and ${\text{20 cm"^3 color(white)(l) "O}}_{2}$.

#### Explanation:

All the substances are gases, so we can use Gay-Lussac's Law of Combining Volumes:

The ratio of the volumes in which gases react and form products is the same as the ratio of their moles.

The equation for the reaction is

${\text{Theor. Vol./cm}}^{3} : \textcolor{w h i t e}{m l} 2 \textcolor{w h i t e}{m m l l} 1 \textcolor{w h i t e}{m m m} 2$
$\textcolor{w h i t e}{m m m m m m m m m} {\text{2CO" + "O"_2 → "2CO}}_{2}$
${\text{Actual Vol./cm}}^{3} : \textcolor{w h i t e}{m} 20 \textcolor{w h i t e}{m m} 30$
$\text{Divide by:} \textcolor{w h i t e}{m m m m m l} 2 \textcolor{w h i t e}{m m l l} 1$
$\text{Moles of reaction:} \textcolor{w h i t e}{m l} 10 \textcolor{w h i t e}{m m} 30$

The coefficients of $\text{CO}$ and ${\text{O}}_{2}$ in the equation tell us that the molecules react in a 2:1 ratio.

A quick way to identify a limiting reactant is to calculate its "moles of reaction".

We divide the number of moles (here they are ${\text{cm}}^{3}$) by their coefficients in the balanced equation.

I did that for you in the lines above.

We see that $\text{CO}$ is the limiting reactant, because it gives the fewest moles of reaction.

Calculate volume of ${\text{CO}}_{2}$ formed

$\text{Vol. of CO"_2 = 20 color(red)(cancel(color(black)("cm"^3color(white)(l) "CO"))) × ("2 cm"^3 color(white)(l)"CO"_2)/(2 color(red)(cancel(color(black)("cm"^3 color(white)(l)"CO")))) = "20 cm"^3color(white)(l) "CO}$

Calculate volume of ${\text{O}}_{2}$ reacted

${\text{Vol of O"_2 color(white)(l)"reacted" = 20 color(red)(cancel(color(black)("cm"^3color(white)(l) "CO"))) × ("1 cm"^3color(white)(l) "O"_2)/(2 color(red)(cancel(color(black)("cm"^3 color(white)(l)"CO")))) = "10 cm"^3color(white)(l) "O}}_{2}$

Calculate the volume of ${\text{O}}_{2}$ remaining

${\text{O"_2color(white)(l) "remaining" = "30 cm"^3 - "10 cm"^3 = "20 cm}}^{3}$