Question

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

Answer 1

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

Explanation:

`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.