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.
