**a) Balanced equation**

#"2CO + O"_2 → "2CO"_2#

**b) i) Volume of oxygen in the air**

#" 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.

#color(white)(mmmmmmm)"2CO" + color(white)(m)"O"_2 → "2CO"_2#

#"Molar ratio:"color(white)(mIll)2color(white)(mmmll)1color(white)(mmm)2#

#V"/cm"^3:"color(white)(mmmll)10color(white)(mmm)21#

#"Divide by:"color(white)(mmm)2color(white)(mmmll)1#

#"Moles rxn:"color(white)(mmm)5color(white)(mmm)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 #"10 cm"^3color(white)(l) "CO"# and #"21 cm"^3color(white)(l) "O"_2# is the same as a mixture containing #"10 mol CO"# and #"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.

#"CO""# is the **limiting reactant** because it gives the fewest "moles of reaction", and #"O"_2# is the **excess reactant**.

#"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#

#"Volume of unused O"_2 = "(21 - 5.0) cm"^3 = "16 cm"^3#

**b) iii) Volume of #"CO"_2# produced**

#"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 #"CO"_2#, unused #"O"_2#, and unused air (mostly nitrogen).

#"Volume = 10 cm"^3 + "16 cm"^3 + "79 cm"^3 = "105 cm"^3#