# Question #31f01

Mar 10, 2017

${C}_{2} {H}_{4} {O}_{2}$ + $2 {O}_{2}$ $\rightarrow$ $2 C {O}_{2}$ + $2 {H}_{2} O$

#### Explanation:

1.) Make a tally sheet of all atoms involved.

${C}_{2} {H}_{4} {O}_{2}$ + ${O}_{2}$ $\rightarrow$ $C {O}_{2}$ + ${H}_{2} O$ (not balanced)

Left Side:
C = 2
H = 4
O = 2 + 2 (DO NOT ADD THIS UP YET)

Right Side:
C = 1
H = 2
O = 2 + 1 (DO NOT ADD THIS UP YET)

2.) Identify the atom that is easiest to balance, in this case, the $C$ atom.

Left Side:
C = 2
H = 4
O = 2 + 2 (DO NOT ADD THIS UP YET)

Right Side:
C = 1 x $\textcolor{red}{2}$ = 2
H = 2
O = 2 + 1 (DO NOT ADD THIS UP YET)

Since the $C$ atom is chemically bonded with the $O$ atom, you would need to also multiply the attached $O$ atom by 2. Hence,

Left Side:
C = 2
H = 4
O = 2 + 2

Right Side:
C = 1 x $\textcolor{red}{2}$ = 2
H = 2
O = (2 x $\textcolor{red}{2}$) + 1

${C}_{2} {H}_{4} {O}_{2}$ + ${O}_{2}$ $\rightarrow$ $\textcolor{red}{2} C {O}_{2}$ + ${H}_{2} O$

3.) Find the next atom that is easy to balance, this time the $H$ atom.

Left Side:
C = 2
H = 4
O = 2 + 2

Right Side:
C = 1 x 2 = 2
H = 2 x $\textcolor{b l u e}{2}$ = 4
O = (2 x 2) + 1

Again, since the $H$ atom is also bonded with another $O$ atom, we apply the same technique as previous.

Left Side:
C = 2
H = 4
O = 2 + 2 = 4

Right Side:
C = 1 x 2 = 2
H = 2 x $\textcolor{b l u e}{2}$ = 4
O = (2 x 2) + (1 x $\textcolor{b l u e}{2}$) = 6

${C}_{2} {H}_{4} {O}_{2}$ + ${O}_{2}$ $\rightarrow$ $\textcolor{red}{2} C {O}_{2}$ + $\textcolor{b l u e}{2} {H}_{2} O$

4.) Notice now that the only atom left unbalanced is the $O$ atoms. You have only 4 atoms on the left side of the equation as opposed to 6 $O$ atoms on the right.

For this, please remember that if given a choice, ALWAYS TRY TO BALANCE USING THE SIMPLER MOLECULE FIRST. In this case, the simpler molecule is the ${O}_{2}$. Thus,

Left Side:
C = 2
H = 4
O = 2 + (2 x $\textcolor{g r e e n}{2}$) = 6

Right Side:
C = 1 x 2 = 2
H = 2 x 2 = 4
O = (2 x 2) + (1 x 2) = 6

${C}_{2} {H}_{4} {O}_{2}$ + $\textcolor{g r e e n}{2} {O}_{2}$ $\rightarrow$ $\textcolor{red}{2} C {O}_{2}$ + $\textcolor{b l u e}{2} {H}_{2} O$