# Question #511ea

Jun 11, 2015

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

#### Explanation:

The unbalanced combustion reaction of hydrogen sulfide, ${H}_{2} S$, looks like this

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

This is a highly exothermic reaction, so the water will be produced as steam. To balance this equation, you need to match the number of atoms present on the reactants' side with the number of atoms present on the products' side.

You have 1 sulfur atom on the reactants' side and 1 on the products' side, so sulfur is balanced.

Notice that you have 2 oxygen atoms on the reactants' side, and 3 on products' side. In order to balance the oxygen atoms, multiply the oxygen molecule by 3, the sulfur dioxide molecule by 2, and the water molecule by 2

${H}_{2} {S}_{\left(g\right)} + \textcolor{red}{3} {O}_{2 \left(g\right)} \to \textcolor{b l u e}{2} S {O}_{2 \left(g\right)} + \textcolor{g r e e n}{2} {H}_{2} {O}_{\left(g\right)}$

The oxygen atoms are now balanced - you have 6 on the lefthand side of the equation and 6 on the righthand side of the equation.

Notice that the sulfur atoms are no longer balanced. To balance the sulfur again, multiply the hydrogen sulfide molecule by 2

$2 {H}_{2} {S}_{\left(g\right)} + \textcolor{red}{3} {O}_{2 \left(g\right)} \to \textcolor{b l u e}{2} S {O}_{2 \left(g\right)} + \textcolor{g r e e n}{2} {H}_{2} {O}_{\left(g\right)}$

The hydrogen atoms are balanced, since you ahve 4 on the lefthand side and *48 on the righthand side of the equation.

All the atoms are balanced, so this is how the balanced chemical equation looks like

$2 {H}_{2} {S}_{\left(g\right)} + \textcolor{red}{3} {O}_{2 \left(g\right)} \to \textcolor{b l u e}{2} S {O}_{2 \left(g\right)} + \textcolor{g r e e n}{2} {H}_{2} {O}_{\left(g\right)}$