Question #31f01

1 Answer
Mar 10, 2017

Answer:

#C_2H_4O_2# + #2O_2# #rarr# #2CO_2# + #2H_2O#

Explanation:

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

#C_2H_4O_2# + #O_2# #rarr# #CO_2# + #H_2O# (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 #color(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 #color(red)2# = 2
H = 2
O = (2 x #color(red)2#) + 1

#C_2H_4O_2# + #O_2# #rarr# #color(red)2CO_2# + #H_2O#

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 #color(blue)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 #color(blue)2# = 4
O = (2 x 2) + (1 x #color(blue)2#) = 6

#C_2H_4O_2# + #O_2# #rarr# #color(red)2CO_2# + #color(blue)2H_2O#

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 #color(green)2#) = 6

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

Final Answer:

#C_2H_4O_2# + #color(green)2O_2# #rarr# #color(red)2CO_2# + #color(blue)2H_2O#

The equation is now balanced.