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

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

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Answer 1 · 2017-02-08T15:21:52

$2 C_{6} H_{6} + 15 O_{2} = 12 C O_{2} + 6 H_{2} O$ $color(red)2C_6H_6+color(red)15O_2 =color(red)12CO_2+color(red)6H_2O$

Explanation:

Let the variables $p, q, r, s$ $color(red)(p, q, r, s)$ represent integers such that:
$XX p \cdot C_{6} H_{6} + q \cdot O_{2} = r \cdot C O_{2} + s \cdot H_{2} O$ $color(white)("XX")color(red)p * C_6H_6 +color(red)q * O_2 = color(red)r * CO_2 +color(red)s * H_2O$

Considering the element $C$ $C$ , we have:
$XX p \cdot 6 = r \cdot 1$ $color(white)("XX")color(red)p * 6 =color(red)r * 1$
$XXXX \to r = 6 p$ $color(white)("XXXX")rarr color(red)r = 6color(red)p$

Considering the element $H$ $H$ , we have:
$XX p \cdot 6 = s \cdot 2$ $color(white)("XX")color(red)p * 6 = color(red)s * 2$
$XXXX \to s = 3 p$ $color(white)("XXXX")rarr color(red)s = 3color(red)p$

Considering the element $O$ $O$ , we have:
$XX q \cdot 2 = r \cdot 2 + s \cdot 1 = 12 p + 3 p = 15 p$ $color(white)("XX")color(red)q * 2 = color(red)r * 2 + color(red)s * 1 = 12color(red)p +3color(red)p = 15color(red)p$
$XXXX \to q = (\frac{15}{2}) p$ $color(white)("XXXX")rarr color(red)q=(15/2)color(red)p$

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

Answer 2 · 2017-02-09T02:34:22

$C_{6}$ $C_6$ $H_{6}$ $H_6$ + $\frac{15}{2}$ $15/2$ $O_{2}$ $O_2$ $\to$ $rarr$ 6 $C O_{2}$ $CO_2$ + 3 $H_{2}$ $H_2$ O

Explanation:

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

$C_{6}$ $C_6$ $H_{6}$ $H_6$ + $O_{2}$ $O_2$ $\to$ $rarr$ $C O_{2}$ $CO_2$ + $H_{2}$ $H_2$ O

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

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

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

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

Right side:
$C$ $C$ = 1 x $6$ $color(red)(6)$ = 6
$H$ $H$ = 2
$O$ $O$ = 2 + 1

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

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

Right side:
$C$ $C$ = 1 x $6$ $color(red)(6)$ = 6
$H$ $H$ = 2
$O$ $O$ = (2 x $6$ $color(red)(6)$ ) + 1

$C_{6}$ $C_6$ $H_{6}$ $H_6$ + $O_{2}$ $O_2$ $\to$ $rarr$ $6$ $color(red)(6)$ $C O_{2}$ $CO_2$ + $H_{2}$ $H_2$ O

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

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

Right side:
$C$ $C$ = 1 x $6$ $color(red)(6)$ = 6
$H$ $H$ = 2 x $3$ $color(green)(3)$ = 6
$O$ $O$ = (2 x $6$ $color(red)(6)$ ) + 1

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

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

Right side:
$C$ $C$ = 1 x $6$ $color(red)(6)$ = 6
$H$ $H$ = 2 x $3$ $color(green)(3)$ = 6
$O$ $O$ = (2 x $6$ $color(red)(6)$ ) + (1 x $3$ $color(green)(3)$ ) = 15

$C_{6}$ $C_6$ $H_{6}$ $H_6$ + $O_{2}$ $O_2$ $\to$ $rarr$ $6$ $color(red)(6)$ $C O_{2}$ $CO_2$ + $3$ $color(green)(3)$ $H_{2}$ $H_2$ O

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

Left side:
$C$ $C$ = 6
$H$ $H$ = 6
$O$ $O$ = 2 x $\frac{15}{2}$ $color(blue)(15/2)$ = 15

Right side:
$C$ $C$ = 1 x $6$ $color(red)(6)$ = 6
$H$ $H$ = 2 x $3$ $color(green)(3)$ = 6
$O$ $O$ = (2 x $6$ $color(red)(6)$ ) + (1 x $3$ $color(green)(3)$ ) = 15

$C_{6}$ $C_6$ $H_{6}$ $H_6$ + $\frac{15}{2}$ $color(blue)(15/2)$ $O_{2}$ $O_2$ $\to$ $rarr$ $6$ $color(red)(6)$ $C O_{2}$ $CO_2$ + $3$ $color(green)(3)$ $H_{2}$ $H_2$ O

The equation is now balanced.

Answer 3 · 2017-02-09T02:34:35

$C_{6}$ $C_6$ $H_{6}$ $H_6$ + $\frac{15}{2}$ $15/2$ $O_{2}$ $O_2$ $\to$ $rarr$ 6 $C O_{2}$ $CO_2$ + 3 $H_{2}$ $H_2$ O

Explanation:

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

$C_{6}$ $C_6$ $H_{6}$ $H_6$ + $O_{2}$ $O_2$ $\to$ $rarr$ $C O_{2}$ $CO_2$ + $H_{2}$ $H_2$ O

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

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

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

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

Right side:
$C$ $C$ = 1 x $6$ $color(red)(6)$ = 6
$H$ $H$ = 2
$O$ $O$ = 2 + 1

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

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

Right side:
$C$ $C$ = 1 x $6$ $color(red)(6)$ = 6
$H$ $H$ = 2
$O$ $O$ = (2 x $6$ $color(red)(6)$ ) + 1

$C_{6}$ $C_6$ $H_{6}$ $H_6$ + $O_{2}$ $O_2$ $\to$ $rarr$ $6$ $color(red)(6)$ $C O_{2}$ $CO_2$ + $H_{2}$ $H_2$ O

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

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

Right side:
$C$ $C$ = 1 x $6$ $color(red)(6)$ = 6
$H$ $H$ = 2 x $3$ $color(green)(3)$ = 6
$O$ $O$ = (2 x $6$ $color(red)(6)$ ) + 1

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

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

Right side:
$C$ $C$ = 1 x $6$ $color(red)(6)$ = 6
$H$ $H$ = 2 x $3$ $color(green)(3)$ = 6
$O$ $O$ = (2 x $6$ $color(red)(6)$ ) + (1 x $3$ $color(green)(3)$ ) = 15

$C_{6}$ $C_6$ $H_{6}$ $H_6$ + $O_{2}$ $O_2$ $\to$ $rarr$ $6$ $color(red)(6)$ $C O_{2}$ $CO_2$ + $3$ $color(green)(3)$ $H_{2}$ $H_2$ O

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

Left side:
$C$ $C$ = 6
$H$ $H$ = 6
$O$ $O$ = 2 x $\frac{15}{2}$ $color(blue)(15/2)$ = 15

Right side:
$C$ $C$ = 1 x $6$ $color(red)(6)$ = 6
$H$ $H$ = 2 x $3$ $color(green)(3)$ = 6
$O$ $O$ = (2 x $6$ $color(red)(6)$ ) + (1 x $3$ $color(green)(3)$ ) = 15

$C_{6}$ $C_6$ $H_{6}$ $H_6$ + $\frac{15}{2}$ $color(blue)(15/2)$ $O_{2}$ $O_2$ $\to$ $rarr$ $6$ $color(red)(6)$ $C O_{2}$ $CO_2$ + $3$ $color(green)(3)$ $H_{2}$ $H_2$ O

The equation is now balanced.

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

[ $C_{6}$ $C_6$ $H_{6}$ $H_6$ + $\frac{15}{2}$ $color(blue)(15/2)$ $O_{2}$ $O_2$ $\to$ $rarr$ $6$ $color(red)(6)$ $C O_{2}$ $CO_2$ + $3$ $color(green)(3)$ $H_{2}$ $H_2$ O] x 2

$2$ $2$ $C_{6}$ $C_6$ $H_{6}$ $H_6$ + $15$ $color(blue)(15)$ $O_{2}$ $O_2$ $\to$ $rarr$ $12$ $color(red)(12)$ $C O_{2}$ $CO_2$ + $6$ $color(green)(6)$ $H_{2}$ $H_2$ O (also acceptable)

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

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