This is a limiting reactant problem.
We know that we will need a balanced equation, molar masses, moles, etc, so let's gather them all in one place now.
#M_text(r):color(white)(mmml)16.04color(white)(ml)32.00color(white)(ml)44.01#
#color(white)(mmmmmm)"CH"_4 + "2O"_2 → "CO"_2 + "2H"_2"O"#
#"Mass/g:" color(white)(mm)23.2color(white)(mll)78.3#
#"Moles:"color(white)(mml) 1.446color(white)(ml)2.447#
#"Divide by:" color(white)(mll)1color(white)(mmm)2#
#"Moles rxn:"color(white)(l)1.446color(white)(ml)1.223#
Identify the limiting reactant
A convenient way to identify the limiting reactant is to calculate its "moles of reaction".
You divide its number of moles by the coefficient in the balanced equation. I did that for you above.
We see that #"O"_2# is the limiting reactant because it gives the fewest moles of reaction.
Calculate the theoretical yield
#"Theor. yield" = 2.447 color(red)(cancel(color(black)("mol O"_2))) × (1 color(red)(cancel(color(black)("mol CO"_2))))/(2 color(red)(cancel(color(black)("mol O"_2)))) × ("44.01 g CO"_2)/(1 color(red)(cancel(color(black)("mol CO"_2)))) = "53.8 g CO"_2#
Calculate the % yield
#"Percent yield" = (52.7 color(red)(cancel(color(black)("g"))))/(53.84 color(red)(cancel(color(black)("g")))) ×100 % = 97.9 %#
