Start with a balanced equation.
#"2C"_6"H"_14 + "19O"_2##rarr##"12CO"_2 + "14H"_2"O"#
Use the balanced equation to determine the mole ratios between #"C"_6"H"_14"# and #"O"_2"#.
#("2 mol C"_6"H"_14)/("19 mol O"_2")# and #("19 mol O"_2)/("2 mol C"_6"H"_14)#
Determine the molar masses of #"C"_6"H"_14# and #"O"_2"#.
#"C"_6"H"_14:##"86.178 g/mol"#
https://www.ncbi.nlm.nih.gov/pccompound?term=C6H14
#"O"_2:##"31.998 g/mol"#
Determine the moles of #"C"_6"H"_14"# by dividing its given mass by its molar mass.
#(11.5 cancel"g C"_6"H"_14)/(86.178cancel"g"/"mol")="0.13344 mol C"_6"H"_14"#
Determine moles of #"O"_2"# by multiplying mol #"C"_6"H"_14"# by the mol ratio that has #"O"_2"# in the numerator.
#0.13344cancel"mol C"_6"H"_14xx(19"mol O"_2)/(2 cancel"mol C"_6"H"_14)="1.2678 mol O"_2"#
Determine the mass of #"O"_2"# that will react with #11.5"g C"_6"H"_14"# by multiplying the moles #"O"_2"# by its molar mass.
#1.2678cancel"mol O"_2xx(31.998"g O"_2)/(1cancel"mol O"_2)="0.0400 g O"_2"# (rounded to three significant figures)
