# What is the balanced equation of C6H6+O2 = CO2+H2O?

Feb 8, 2017

$\textcolor{red}{2} {C}_{6} {H}_{6} + \textcolor{red}{15} {O}_{2} = \textcolor{red}{12} C {O}_{2} + \textcolor{red}{6} {H}_{2} O$

#### Explanation:

Let the variables $\textcolor{red}{p , q , r , s}$ represent integers such that:
$\textcolor{w h i t e}{\text{XX}} \textcolor{red}{p} \cdot {C}_{6} {H}_{6} + \textcolor{red}{q} \cdot {O}_{2} = \textcolor{red}{r} \cdot C {O}_{2} + \textcolor{red}{s} \cdot {H}_{2} O$

Considering the element $C$, we have:
$\textcolor{w h i t e}{\text{XX}} \textcolor{red}{p} \cdot 6 = \textcolor{red}{r} \cdot 1$
$\textcolor{w h i t e}{\text{XXXX}} \rightarrow \textcolor{red}{r} = 6 \textcolor{red}{p}$

Considering the element $H$, we have:
$\textcolor{w h i t e}{\text{XX}} \textcolor{red}{p} \cdot 6 = \textcolor{red}{s} \cdot 2$
$\textcolor{w h i t e}{\text{XXXX}} \rightarrow \textcolor{red}{s} = 3 \textcolor{red}{p}$

Considering the element $O$, we have:
$\textcolor{w h i t e}{\text{XX}} \textcolor{red}{q} \cdot 2 = \textcolor{red}{r} \cdot 2 + \textcolor{red}{s} \cdot 1 = 12 \textcolor{red}{p} + 3 \textcolor{red}{p} = 15 \textcolor{red}{p}$
$\textcolor{w h i t e}{\text{XXXX}} \rightarrow \textcolor{red}{q} = \left(\frac{15}{2}\right) \textcolor{red}{p}$

The smallest value of $\textcolor{red}{p} > 0$ for which $\textcolor{red}{q}$ is an integer is:
$\textcolor{w h i t e}{\text{XX}} \textcolor{red}{p} = \textcolor{red}{2}$
color(white)("XXXX")rarr {color(red)q=color(red)15; color(red)r=color(red)12; color(red)s=color(red)6}

Feb 9, 2017

${C}_{6}$${H}_{6}$ + $\frac{15}{2}$ ${O}_{2}$ $\rightarrow$ 6$C {O}_{2}$+ 3${H}_{2}$O

#### Explanation:

1.) Create a tally sheet of atoms involved in the reaction.

${C}_{6}$${H}_{6}$ + ${O}_{2}$ $\rightarrow$ $C {O}_{2}$+ ${H}_{2}$O

Left side:
$C$ = 6
$H$ = 6
$O$ = 2

Right side:
$C$ = 1
$H$ = 2
$O$ = 2 + 1 ( DO NOT ADD IT UP YET )

2.) Find the atoms that are easiest to balance. In this case, the $C$ atom.

Left side:
$C$ = 6
$H$ = 6
$O$ = 2

Right side:
$C$ = 1 x $\textcolor{red}{6}$ = 6
$H$ = 2
$O$ = 2 + 1

3.) Remember that in the equation, the $C$ atom is a part of a substance. Therefore, you have to multiply the attached $O$ atom with the factor as well.

Left side:
$C$ = 6
$H$ = 6
$O$ = 2

Right side:
$C$ = 1 x $\textcolor{red}{6}$ = 6
$H$ = 2
$O$ = (2 x $\textcolor{red}{6}$) + 1

${C}_{6}$${H}_{6}$ + ${O}_{2}$ $\rightarrow$ $\textcolor{red}{6}$$C {O}_{2}$+ ${H}_{2}$O

4.) Find the next atom to balance. In this case, the $H$ atom.

Left side:
$C$ = 6
$H$ = 6
$O$ = 2

Right side:
$C$ = 1 x $\textcolor{red}{6}$ = 6
$H$ = 2 x $\textcolor{g r e e n}{3}$ = 6
$O$ = (2 x $\textcolor{red}{6}$) + 1

Again, the $H$ atom is chemically bonded to another $O$ atom. Thus,

Left side:
$C$ = 6
$H$ = 6
$O$ = 2

Right side:
$C$ = 1 x $\textcolor{red}{6}$ = 6
$H$ = 2 x $\textcolor{g r e e n}{3}$ = 6
$O$ = (2 x $\textcolor{red}{6}$) + (1 x $\textcolor{g r e e n}{3}$) = 15

${C}_{6}$${H}_{6}$ + ${O}_{2}$ $\rightarrow$ $\textcolor{red}{6}$$C {O}_{2}$+ $\textcolor{g r e e n}{3}$${H}_{2}$O

5.) Balance out the remaining $O$ atoms. Since the $O$ atoms on the right side is an odd number, you can use your knowledge of fractions to get the complete balanced equation.

Left side:
$C$ = 6
$H$ = 6
$O$ = 2 x $\textcolor{b l u e}{\frac{15}{2}}$ = 15

Right side:
$C$ = 1 x $\textcolor{red}{6}$ = 6
$H$ = 2 x $\textcolor{g r e e n}{3}$ = 6
$O$ = (2 x $\textcolor{red}{6}$) + (1 x $\textcolor{g r e e n}{3}$) = 15

${C}_{6}$${H}_{6}$ + $\textcolor{b l u e}{\frac{15}{2}}$${O}_{2}$ $\rightarrow$ $\textcolor{red}{6}$$C {O}_{2}$+ $\textcolor{g r e e n}{3}$${H}_{2}$O

The equation is now balanced.

Feb 9, 2017

${C}_{6}$${H}_{6}$ + $\frac{15}{2}$ ${O}_{2}$ $\rightarrow$ 6$C {O}_{2}$+ 3${H}_{2}$O

#### Explanation:

1.) Create a tally sheet of atoms involved in the reaction.

${C}_{6}$${H}_{6}$ + ${O}_{2}$ $\rightarrow$ $C {O}_{2}$+ ${H}_{2}$O

Left side:
$C$ = 6
$H$ = 6
$O$ = 2

Right side:
$C$ = 1
$H$ = 2
$O$ = 2 + 1 ( DO NOT ADD IT UP YET )

2.) Find the atoms that are easiest to balance. In this case, the $C$ atom.

Left side:
$C$ = 6
$H$ = 6
$O$ = 2

Right side:
$C$ = 1 x $\textcolor{red}{6}$ = 6
$H$ = 2
$O$ = 2 + 1

3.) Remember that in the equation, the $C$ atom is a part of a substance. Therefore, you have to multiply the attached $O$ atom with the factor as well.

Left side:
$C$ = 6
$H$ = 6
$O$ = 2

Right side:
$C$ = 1 x $\textcolor{red}{6}$ = 6
$H$ = 2
$O$ = (2 x $\textcolor{red}{6}$) + 1

${C}_{6}$${H}_{6}$ + ${O}_{2}$ $\rightarrow$ $\textcolor{red}{6}$$C {O}_{2}$+ ${H}_{2}$O

4.) Find the next atom to balance. In this case, the $H$ atom.

Left side:
$C$ = 6
$H$ = 6
$O$ = 2

Right side:
$C$ = 1 x $\textcolor{red}{6}$ = 6
$H$ = 2 x $\textcolor{g r e e n}{3}$ = 6
$O$ = (2 x $\textcolor{red}{6}$) + 1

Again, the $H$ atom is chemically bonded to another $O$ atom. Thus,

Left side:
$C$ = 6
$H$ = 6
$O$ = 2

Right side:
$C$ = 1 x $\textcolor{red}{6}$ = 6
$H$ = 2 x $\textcolor{g r e e n}{3}$ = 6
$O$ = (2 x $\textcolor{red}{6}$) + (1 x $\textcolor{g r e e n}{3}$) = 15

${C}_{6}$${H}_{6}$ + ${O}_{2}$ $\rightarrow$ $\textcolor{red}{6}$$C {O}_{2}$+ $\textcolor{g r e e n}{3}$${H}_{2}$O

5.) Balance out the remaining $O$ atoms. Since the $O$ atoms on the right side is an odd number, you can use your knowledge of fractions to get the complete balanced equation.

Left side:
$C$ = 6
$H$ = 6
$O$ = 2 x $\textcolor{b l u e}{\frac{15}{2}}$ = 15

Right side:
$C$ = 1 x $\textcolor{red}{6}$ = 6
$H$ = 2 x $\textcolor{g r e e n}{3}$ = 6
$O$ = (2 x $\textcolor{red}{6}$) + (1 x $\textcolor{g r e e n}{3}$) = 15

${C}_{6}$${H}_{6}$ + $\textcolor{b l u e}{\frac{15}{2}}$${O}_{2}$ $\rightarrow$ $\textcolor{red}{6}$$C {O}_{2}$+ $\textcolor{g r e e n}{3}$${H}_{2}$O

The equation is now balanced.

If you don't like fractions, you can always multiply the whole chemical equation by the denominator.

[${C}_{6}$${H}_{6}$ + $\textcolor{b l u e}{\frac{15}{2}}$${O}_{2}$ $\rightarrow$ $\textcolor{red}{6}$$C {O}_{2}$+ $\textcolor{g r e e n}{3}$${H}_{2}$O] x 2

$2$ ${C}_{6}$${H}_{6}$ + $\textcolor{b l u e}{15}$${O}_{2}$ $\rightarrow$ $\textcolor{red}{12}$ $C {O}_{2}$+ $\textcolor{g r e e n}{6}$ ${H}_{2}$O (also acceptable)