Question

Question #31f01

Answer 1

`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.