How would you balance: $C_{12} H_{22} O_{11} + H_{2} O \to C_{2} H_{5} O H + C O_{2}$ $C_12H_22O_11 + H_2O -> C_2H_5OH + CO_2$ ?

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How would you balance: $C_{12} H_{22} O_{11} + H_{2} O \to C_{2} H_{5} O H + C O_{2}$ $C_12H_22O_11 + H_2O -> C_2H_5OH + CO_2$ ?

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Answer 1 · 2016-01-10T14:47:54

$C_{12} H_{22} O_{11} + H_{2} O \to 4 C_{2} H_{5} O H + 4 C O_{2}$ $C_12 H_22 O_11 + H_2 O -> 4 C_2 H_5 OH + 4 CO_2$

Explanation:

$C_{12} H_{22} O_{11} + H_{2} O \to C_{2} H_{5} O H + C O_{2}$ $C_12 H_22 O_11 + H_2 O -> C_2 H_5 OH + CO_2$

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

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

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How would you balance: $C_{12} H_{22} O_{11} + H_{2} O \to C_{2} H_{5} O H + C O_{2}$ $C_12H_22O_11 + H_2O -> C_2H_5OH + CO_2$ ?

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How would you balance: C12H22O11+H2O→C2H5OH+CO2C_12H_22O_11 + H_2O -> C_2H_5OH + CO_2?

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How would you balance: $C_{12} H_{22} O_{11} + H_{2} O \to C_{2} H_{5} O H + C O_{2}$ $C_12H_22O_11 + H_2O -> C_2H_5OH + CO_2$ ?