Question #31f01

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Question #31f01

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Answer 1 · 2017-03-10T09:07:29

$C_{2} H_{4} O_{2}$ $C_2H_4O_2$ + $2 O_{2}$ $2O_2$ $\to$ $rarr$ $2 C O_{2}$ $2CO_2$ + $2 H_{2} O$ $2H_2O$

Explanation:

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

$C_{2} H_{4} O_{2}$ $C_2H_4O_2$ + $O_{2}$ $O_2$ $\to$ $rarr$ $C O_{2}$ $CO_2$ + $H_{2} O$ $H_2O$ (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 $C$ $C$ atom.

Left Side:
C = 2
H = 4
O = 2 + 2 (DO NOT ADD THIS UP YET)

Right Side:
C = 1 x $2$ $color(red)2$ = 2
H = 2
O = 2 + 1 (DO NOT ADD THIS UP YET)

Since the $C$ $C$ atom is chemically bonded with the $O$ $O$ atom, you would need to also multiply the attached $O$ $O$ atom by 2. Hence,

Left Side:
C = 2
H = 4
O = 2 + 2

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

$C_{2} H_{4} O_{2}$ $C_2H_4O_2$ + $O_{2}$ $O_2$ $\to$ $rarr$ $2 C O_{2}$ $color(red)2CO_2$ + $H_{2} O$ $H_2O$

3.) Find the next atom that is easy to balance, this time the $H$ $H$ atom.

Left Side:
C = 2
H = 4
O = 2 + 2

Right Side:
C = 1 x 2 = 2
H = 2 x $2$ $color(blue)2$ = 4
O = (2 x 2) + 1

Again, since the $H$ $H$ atom is also bonded with another $O$ $O$ 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 $2$ $color(blue)2$ = 4
O = (2 x 2) + (1 x $2$ $color(blue)2$ ) = 6

$C_{2} H_{4} O_{2}$ $C_2H_4O_2$ + $O_{2}$ $O_2$ $\to$ $rarr$ $2 C O_{2}$ $color(red)2CO_2$ + $2 H_{2} O$ $color(blue)2H_2O$

4.) Notice now that the only atom left unbalanced is the $O$ $O$ atoms. You have only 4 atoms on the left side of the equation as opposed to 6 $O$ $O$ 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 $O_{2}$ $O_2$ . Thus,

Left Side:
C = 2
H = 4
O = 2 + (2 x $2$ $color(green)2$ ) = 6

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

Final Answer:

$C_{2} H_{4} O_{2}$ $C_2H_4O_2$ + $2 O_{2}$ $color(green)2O_2$ $\to$ $rarr$ $2 C O_{2}$ $color(red)2CO_2$ + $2 H_{2} O$ $color(blue)2H_2O$

The equation is now balanced.

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Question #31f01

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