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

${C}_{12} {H}_{22} {O}_{11} + {H}_{2} O \to 4 {C}_{2} {H}_{5} O H + 4 C {O}_{2}$
${C}_{12} {H}_{22} {O}_{11} + {H}_{2} O \to {C}_{2} {H}_{5} O H + C {O}_{2}$
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.