1. Write the balanced equation.
#"2C"_6"H"_14 + "19O"_2 → "12CO"_2 + "14H"_2"O"#
2. Calculate the moles of #"C"_6"H"_14#.
#"Moles of C"_6"H"_14 = 84.27 color(red)(cancel(color(black)("g C"_6"H"_14))) × ("1 mol C"_6"H"_14)/(86.17 color(red)(cancel(color(black)("g C"_6"H"_14)))) = "0.977 95 mol C"_6"H"_14"#
3. Calculate the moles of #"H"_2"O"#.
#"Moles of H"_2"O" = "0.977 95" color(red)(cancel(color(black)("mol C"_6"H"_14))) × ("14 mol H"_2"O")/(2 color(red)(cancel(color(black)("mol C"_6"H"_14)))) = "6.8457 mol H"_2"O"#
4. Calculate the mass of #"H"_2"O"#.
#"Mass of H"_2"O" = 6.8457 color(red)(cancel(color(black)("mol H"_2"O"))) × ("18.02 g H"_2)/(1 color(red)(cancel(color(black)("mol H"_2"O")))) = "123.4 g H"_2"O"#