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

3 Answers
Alan P.
Feb 8, 2017

#color(red)2C_6H_6+color(red)15O_2 =color(red)12CO_2+color(red)6H_2O#

Explanation:

Let the variables #color(red)(p, q, r, s)# represent integers such that:
#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#, we have:
#color(white)("XX")color(red)p * 6 =color(red)r * 1#
#color(white)("XXXX")rarr color(red)r = 6color(red)p#

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

Considering the element #O#, we have:
#color(white)("XX")color(red)q * 2 = color(red)r * 2 + color(red)s * 1 = 12color(red)p +3color(red)p = 15color(red)p#
#color(white)("XXXX")rarr color(red)q=(15/2)color(red)p#

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

Nikka C.
Feb 9, 2017

#C_6##H_6# + #15/2# #O_2# #rarr# 6#CO_2#+ 3#H_2#O

Explanation:

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

#C_6##H_6# + #O_2# #rarr# #CO_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 #color(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 #color(red)(6)# = 6
#H# = 2
#O# = (2 x #color(red)(6)#) + 1

#C_6##H_6# + #O_2# #rarr# #color(red)(6)##CO_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 #color(red)(6)# = 6
#H# = 2 x #color(green)(3)# = 6
#O# = (2 x #color(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 #color(red)(6)# = 6
#H# = 2 x #color(green)(3)# = 6
#O# = (2 x #color(red)(6)#) + (1 x #color(green)(3)#) = 15

#C_6##H_6# + #O_2# #rarr# #color(red)(6)##CO_2#+ #color(green)(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 #color(blue)(15/2)# = 15

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

#C_6##H_6# + #color(blue)(15/2)##O_2# #rarr# #color(red)(6)##CO_2#+ #color(green)(3)##H_2#O

The equation is now balanced.

Nikka C.
Feb 9, 2017

#C_6##H_6# + #15/2# #O_2# #rarr# 6#CO_2#+ 3#H_2#O

Explanation:

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

#C_6##H_6# + #O_2# #rarr# #CO_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 #color(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 #color(red)(6)# = 6
#H# = 2
#O# = (2 x #color(red)(6)#) + 1

#C_6##H_6# + #O_2# #rarr# #color(red)(6)##CO_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 #color(red)(6)# = 6
#H# = 2 x #color(green)(3)# = 6
#O# = (2 x #color(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 #color(red)(6)# = 6
#H# = 2 x #color(green)(3)# = 6
#O# = (2 x #color(red)(6)#) + (1 x #color(green)(3)#) = 15

#C_6##H_6# + #O_2# #rarr# #color(red)(6)##CO_2#+ #color(green)(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 #color(blue)(15/2)# = 15

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

#C_6##H_6# + #color(blue)(15/2)##O_2# #rarr# #color(red)(6)##CO_2#+ #color(green)(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# + #color(blue)(15/2)##O_2# #rarr# #color(red)(6)##CO_2#+ #color(green)(3)##H_2#O] x 2

#2# #C_6##H_6# + #color(blue)(15)##O_2# #rarr# #color(red)(12)# #CO_2#+ #color(green)(6)# #H_2#O (also acceptable)

Related questions
See all questions in Balancing Chemical Equations
Impact of this question
173660 views around the world
You can reuse this answer
Creative Commons License