We have the masses of two reactants, so this is a limiting reactant problem.
We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.
Step 1. Gather all the information in one place with molar masses above the formulas and everything else below them.
#M_text(r):color(white)(mmmm)168.32color(white)(mll)32.00#
#color(white)(mmmmmm)"C"_12"H"_24 + "18O"_2 → "12CO"_2 + "12H"_2"O"#
#"Mass/g":color(white)(mml)200color(white)(mmm)12#
Step 2. Calculate the moles of each reactant
#"Moles of C"_12"H"_24 = 200 color(red)(cancel(color(black)("g C"_12"H"_24))) × ("1 mol C"_12"H"_24)/(168.32 color(red)(cancel(color(black)("g C"_12"H"_24)))) = "1.188 mol C"_12"H"_24#
#"Moles of O"_2 = 10 color(red)(cancel(color(black)("g O"_2))) × ("1 mol O"_2)/(32.00 color(red)(cancel(color(black)("g O"_2)))) = "0.312 mol O"_2#
Step 3. Identify the limiting reactant
Calculate the moles of #"CO"_2# we can obtain from each reactant.
From #"C"_12"H"_24#:
#"Moles of CO"_2 = 1.188 color(red)(cancel(color(black)("mol C"_12"H"_24))) × "12 mol CO"_2/(1 color(red)(cancel(color(black)("mol C"_12"H"_24)))) = "14.25 mol CO"_2#
From #"O"_2#:
#"Moles of CO"_2 = 0.312color(red)(cancel(color(black)("mol O"_2))) × "12 mol CO"_2/(18 color(red)(cancel(color(black)("mol O"_2)))) = "0.208 mol CO"_2#
#"O"2# is the limiting reactant because it gives the smaller amount of #"CO"_2#.
Step 4. Calculate the volume of #"CO"_2#
STP is 0 °C and 1 bar.
The molar volume of an ideal gas at STP is 22.71 L. Thus,
#"Volume of CO"_2 = "0.208 mol CO"_2 × "22.71 L CO"_2/(1 "mol CO"_2) = "4.7 L"#
Note: The answer can have only two significant figures because that is all you gave for the mass of oxygen.
