This is a limiting reactant problem.
Step 1. Write the balanced equation.
#"C"_3"H"_8 + "5O"_2 → "3CO"_2 + "4H"_2"O"#
2. Identify the limiting reactant
One way to identify the limiting reactant is to calculate the number of molecules of product each reactant will give.
From #"C"_3"H"_8#:
#3 color(red)(cancel(color(black)("molecules C"_3"H"_8))) × ("3 molecules CO"_2)/(1 color(red)(cancel(color(black)("molecule C"_3"H"_8)))) → "9 molecules CO"_2#
From #"O"_2#:
#10 color(red)(cancel(color(black)("molecules O"_2))) × ("3 molecules CO"_2)/(5 color(red)(cancel(color(black)("molecules O"_2)))) → "6 molecules CO"_2#
#"O"_2# is the limiting reactant because it gives fewer molecules of #"CO"_2#.
Step 3. Calculate the molecules of #"H"_2"O"#
#"Molecules of H"_2"O" = 10 color(red)(cancel(color(black)("molecules O"_2))) × ("4 molecules H"_2"O")/(5 color(red)(cancel(color(black)("molecules O"_2)))) = "8 molecules H"_2"O"#
Step 4. Calculate the molecules of #"C"_3"H"_8# used up
#"Molecules of C"_3"H"_8 color(white)(l)"reacted" = 10 color(red)(cancel(color(black)("molecules O"_2))) × ("1 molecule C"_3"H"_8)/(5 color(red)(cancel(color(black)("molecules O"_2)))) = "2 molecules C"_3"H"_8#
Step 5. Calculate the molecules of #"CO"_2# in excess
#"Excess CO"_2 = "3 molecules CO"_2 - "2 molecules CO"_2 = "1 molecule CO"_2#
When the reaction goes to completion, we will have 1 molecule of #"C"_3"H"_8#, 0 molecules of #"O"_2#, 6 molecules of #"CO"_2#, and 8 molecules of #"H"_2"O"#.
Check:
#color(white)(m)"3C"_3"H"_8 + "10O"_2 = "1C"_3"H"_8 + "6CO"_2 + "8H"_2"O"#
#"9C + 24H + 20O = 9C + 24H + 20O"#
It checks! We have the same number of atoms before and after the reaction.
