# Question #79600

Apr 11, 2016

Sulphur dioxide gas reacts with hydrogen sulphide to give rise to sulphur and water, not oxygen.

#### Explanation:

A little drop of water activates the reaction between the two gases.

$S {O}_{2} \left(g\right) + {H}_{2} S \left(g\right) \to S \left(s\right) + {H}_{2} O \left(l\right)$

Despite it is a redox, the equation can be easily balanced by trials and errors.
Having two oxygen atoms in the reactants, we need four hydrogen atoms, that is two ${H}_{2} S$ molecules, and we obtain three suphur atoms together with two water molecules:

$S {O}_{2} \left(g\right) + 2 {H}_{2} S \left(g\right) \to 3 S \left(s\right) + 2 {H}_{2} O \left(l\right)$

Taking in account that sulphur is octaatomic and the least common multiple between 3 and 8 is 24, we must multiply every coefficient by eight to have the very complete equation:

$8 S {O}_{2} \left(g\right) + 16 {H}_{2} S \left(g\right) \to 3 {S}_{8} \left(s\right) + 16 {H}_{2} O \left(l\right)$