# Question #06c9b

Jun 17, 2017

Warning! Long Answer. a) ${\text{2CO + O"_2 → "2CO}}_{2}$; b) i) ${\text{21 cm}}^{3}$; ii) ${\text{16 cm"^3color(white)(l) "O}}_{2}$; iii) ${\text{10 cm}}^{3}$; iv) ${\text{105 cm}}^{3}$

#### Explanation:

a) Balanced equation

${\text{2CO + O"_2 → "2CO}}_{2}$

b) i) Volume of oxygen in the air

${\text{ Volume of O"_2 = 100 color(red)(cancel(color(black)("cm"^3 color(white)(l)"air"))) × ("21 cm"^3color(white)(l) "O"_2)/(100 color(red)(cancel(color(black)("cm"^3color(white)(l) "air")))) = "21 cm"^3 color(white)(l)"O}}_{2}$

b) ii) Volume of gas unused

This is a limiting reactant problem because we are given the volumes of two reactants.

We know that we will need a balanced equation, volumes, and moles, so let's gather all the information in one place.

$\textcolor{w h i t e}{m m m m m m m} {\text{2CO" + color(white)(m)"O"_2 → "2CO}}_{2}$
$\text{Molar ratio:} \textcolor{w h i t e}{m I l l} 2 \textcolor{w h i t e}{m m m l l} 1 \textcolor{w h i t e}{m m m} 2$
$V \text{/cm"^3:} \textcolor{w h i t e}{m m m l l} 10 \textcolor{w h i t e}{m m m} 21$
$\text{Divide by:} \textcolor{w h i t e}{m m m} 2 \textcolor{w h i t e}{m m m l l} 1$
$\text{Moles rxn:} \textcolor{w h i t e}{m m m} 5 \textcolor{w h i t e}{m m m} 21$

We know from Gay-Lussac's Law of Combining Volumes that the ratio of the volumes is the same as the ratio of the moles.

Thus, a mixture containing $\text{10 cm"^3color(white)(l) "CO}$ and ${\text{21 cm"^3color(white)(l) "O}}_{2}$ is the same as a mixture containing $\text{10 mol CO}$ and ${\text{21 mol O}}_{2}$.

This means that we can treat the volumes as if they were moles.

An easy way to identify the limiting reactant is to calculate the "moles of reaction" each will produce.

You simply divide the number of moles by the corresponding coefficient in the balanced equation.

I did that for you in the chart above.

$\text{CO}$ is the limiting reactant because it gives the fewest "moles of reaction", and ${\text{O}}_{2}$ is the excess reactant.

${\text{Volume of O"_2 color(white)(l)"used" = "10" color(red)(cancel(color(black)("cm"^3color(white)(l) "CO"))) × ("1 cm"^3color(white)(l) "O"_2)/(2color(red)(cancel(color(black)("cm"^3color(white)(l) "CO")))) = "5.0 cm"^3 color(white)(l)"O}}_{2}$

${\text{Volume of unused O"_2 = "(21 - 5.0) cm"^3 = "16 cm}}^{3}$

b) iii) Volume of ${\text{CO}}_{2}$ produced

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

b) iv) Volume of residual gases

The gases at the end of the reaction are ${\text{CO}}_{2}$, unused ${\text{O}}_{2}$, and unused air (mostly nitrogen).

${\text{Volume = 10 cm"^3 + "16 cm"^3 + "79 cm"^3 = "105 cm}}^{3}$