Is the equation $2 C_{8} H_{8} + 25 O_{2} \to 8 C O_{2} + 18 H_{2} O$ $2C_8H_8 + 25O_2 -> 8CO_2 + 18H_2O$ completely balanced?

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Is the equation $2 C_{8} H_{8} + 25 O_{2} \to 8 C O_{2} + 18 H_{2} O$ $2C_8H_8 + 25O_2 -> 8CO_2 + 18H_2O$ completely balanced?

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Answer 1 · 2016-06-06T03:03:39

Count on the number of atoms on each side of the equation.

On the left hand side (or left of the arrow),
C = 2 $\times$ $xx$ 8 = 16 atoms
H = 2 $\times$ $xx$ 8 = 16 atoms
O = 25 $\times$ $xx$ 2 = 50 atoms

On the right hand side,
C = 8 $\times$ $xx$ 1 = 8 atoms
H = 18 $\times$ $xx$ 2 = 36 atoms
O = 8 $\times$ $xx$ 2 $+$ $+$ 18 $\times$ $xx$ 1 = 34 atoms

No, it's not balanced. The number of atoms has to be the same on both sides.

To balance Carbon together on both sides, find the lowest common multiple (LCM) of 16 and 8, therefore its 16, so you need to balance them together to have 16 atoms of C on both sides. Change $8 C O_{2}$ $8CO_2$ to $16 C O_{2}$ $16CO_2$ .
$2 C_{8} H_{8} + 25 O_{2} \Rightarrow 16 C O_{2} + 18 H_{2} O$ $2C_8H_8 + 25O_2 => 16CO_2 + 18H_2O$ .
Note that this is still unbalanced.

Do the same thing for the rest and remember to count up the number of atoms on both sides.

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Is the equation $2 C_{8} H_{8} + 25 O_{2} \to 8 C O_{2} + 18 H_{2} O$ $2C_8H_8 + 25O_2 -> 8CO_2 + 18H_2O$ completely balanced?

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Is the equation $2 C_{8} H_{8} + 25 O_{2} \to 8 C O_{2} + 18 H_{2} O$ $2C_8H_8 + 25O_2 -> 8CO_2 + 18H_2O$ completely balanced?