How do you balance #C_3H_6 + O_2 -> CO + H_2O#?

2 Answers

Answer:

#C_3H_6+3O_2->3CO+3H_2O#

Explanation:

To balance, we add molecules such as to ensure that the number of atoms of each element is the same on each side of the equation.

This equation represents the incomplete combustion of propane as insufficient oxygen must have been present since carbon monoxide was form instead of carbon dioxide which would form if there was sufficient oxygen to ensure complete combustion.

Jan 29, 2016

Answer:

#C_3H_6 + 3O_2 rarr 3CO + 3H_2O#

Explanation:

We have three carbons on the reagents' side and only one on the products' side, so we need to balance that by making that equal.

#C_3H_6 + O_2 rarr 3CO + H_2O#

We have six hydrogens on the reagents' side and only 2 on the products' side so we need to balance that by making them equal.

#C_3H_6 + O_2 rarr 3CO + 3H_2O#

We have 2 oxygens on the reagents' side and 6 on the products' side, so we need to balance that by making them equal, now, since the reagent's side has only one substance with oxygen it'll be easier to make that equal, so,

#C_3H_6 + 3O_2 rarr 3CO + 3H_2O#

Do note that if you had equalized the first time around by

#1/3C_3H_6 + O_2 rarr CO + H_2O#

It'd be already balanced, but it's considered good practice to balance equations by having integer coefficients, which we could have solved by multiplying everything by 3

#C_3H_6 + 3O_2 rarr 3CO + 3H_2O#