How would you balance: $C_12H_22O_11 + H_2O -> C_2H_5OH + CO_2$ ?

Question

2743 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-10T14:47:54

$C_12 H_22 O_11 + H_2 O -> 4 C_2 H_5 OH + 4 CO_2$

$C_12 H_22 O_11 + H_2 O -> C_2 H_5 OH + CO_2$

Counting the occurrences of $C$ , $H$ and $O$ on the left hand side we have half as many $C$ atoms ( $12$ ) as $H$ or $O$ ( $24$ ).

The right hand side already has the same proportions of $C$ , $H$ and $O$ , so we just need the right multiplier for the right hand side.

How would you balance: C_12H_22O_11 + H_2O -> C_2H_5OH + CO_2C_12H_22O_11 + H_2O -> C_2H_5OH + CO_2?