Start with a balanced equation.
$"2C"_6"H"_14 + "19O"_2$$rarr$$"12CO"_2 + "14H"_2"O"$
Use the balanced equation to determine the 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:$$"86.178 g/mol"$
$"O"_2:$$"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)
