A #600*cm^3# volume of ethane gas is completely combusted with #2500*cm^3# of dioxygen gas to give carbon dioxide and water. What is the reagent in excess, and what volume would #CO_2(g)# occupy after the reaction....?

We are given the amounts of two reactants, so this is a limiting reactant problem.

We know that we will need a balanced equation with volumes of the compounds involved.

1. Gather all the information in one place.

#color(white)(mmmmmmll) "2C"_2"H"_6 + color(white)(m)"7O"_2 → "4CO"_2 + "6H"_2"O"#
#V"/cm"^3:color(white)(mmml)600color(white)(mmlll)2500#
#"Divide by:"color(white)(mmm)2color(white)(mmmml)7#
#"Moles rxn:"color(white)(mml)300color(white)(mmml)357#

The molar quantities of the gases are directly proportional to the volumes, so we can use volumes as a measure of moles.

2. Identify the limiting reactant

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

You divide the moles of each reactant by its corresponding coefficient in the balanced equation.

I did that for you in the table above.

#"C"_2"H"_6# is the limiting reactant because it gives the fewer moles of reaction.

3. Calculate the theoretical volume of #"CO"_2#

#"Theoretical yield" = 600 color(red)(cancel(color(black)("cm"^3 color(white)(l)"C"_2"H"_6))) × ("4 cm"^3color(white)(l) "CO"_2)/(2 color(red)(cancel(color(black)("cm"^3color(white)(l) "C"_2"H"_6)))) = "1200 cm"^3 color(white)(l)"CO"_2#

4. Calculate the volume of #"O"_2# used

#"Volume of O"_2 = 600 color(red)(cancel(color(black)("cm"^3 color(white)(l)"C"_2"H"_6))) × ("7 cm"^3color(white)(l) "O"_2)/(2 color(red)(cancel(color(black)("cm"^3 color(white)(l)"C"_2"H"_6)))) = "2100 cm"^3color(white)(l) "O"_2#

5. Calculate the volume of #"O"_2# remaining

#"Volume remaining" = "original volume - volume used" = "2500 cm"^3 - "2100 cm"^3 = "400 cm"^3#