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

1 Answer
Oct 8, 2016

Answer:

There will be 1 molecule of #"C"_3"H"_8#, 0 molecules of #"O"_2#, 6 molecules of #"CO"_2#, and 8 molecules of #"H"_2"O"#.

Explanation:

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.