# How many molecules of each substance are present after the combustion of 3 molecules of propane with 10 molecules of oxygen?

Oct 8, 2016

There will be 1 molecule of ${\text{C"_3"H}}_{8}$, 0 molecules of ${\text{O}}_{2}$, 6 molecules of ${\text{CO}}_{2}$, and 8 molecules of $\text{H"_2"O}$.

#### Explanation:

This is a limiting reactant problem.

Step 1. Write the balanced equation.

$\text{C"_3"H"_8 + "5O"_2 → "3CO"_2 + "4H"_2"O}$

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 ${\text{C"_3"H}}_{8}$:

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 ${\text{O}}_{2}$:

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

${\text{O}}_{2}$ is the limiting reactant because it gives fewer molecules of ${\text{CO}}_{2}$.

Step 3. Calculate the molecules of $\text{H"_2"O}$

$\text{Molecules of H"_2"O" = 10 color(red)(cancel(color(black)("molecules O"_2))) × ("4 molecules H"_2"O")/(5 color(red)(cancel(color(black)("molecules O"_2)))) = "8 molecules H"_2"O}$

Step 4. Calculate the molecules of ${\text{C"_3"H}}_{8}$ used up

${\text{Molecules of C"_3"H"_8 color(white)(l)"reacted" = 10 color(red)(cancel(color(black)("molecules O"_2))) × ("1 molecule C"_3"H"_8)/(5 color(red)(cancel(color(black)("molecules O"_2)))) = "2 molecules C"_3"H}}_{8}$

Step 5. Calculate the molecules of ${\text{CO}}_{2}$ in excess

${\text{Excess CO"_2 = "3 molecules CO"_2 - "2 molecules CO"_2 = "1 molecule CO}}_{2}$

When the reaction goes to completion, we will have 1 molecule of ${\text{C"_3"H}}_{8}$, 0 molecules of ${\text{O}}_{2}$, 6 molecules of ${\text{CO}}_{2}$, and 8 molecules of $\text{H"_2"O}$.

Check:

$\textcolor{w h i t e}{m} \text{3C"_3"H"_8 + "10O"_2 = "1C"_3"H"_8 + "6CO"_2 + "8H"_2"O}$

$\text{9C + 24H + 20O = 9C + 24H + 20O}$

It checks! We have the same number of atoms before and after the reaction.