Translate the following statement into chemical equation and balance it:
Hydrogen sulphide gas burns in air to give water and sulphur dioxide.

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Solution

$2 H_{2} S_{(g)} + 3 O_{2 (g)} + 2 H_{2} O_{(g)}$
Explanation:
The unbalanced combustion reaction of hydrogen sulfide, $H_{2} S$ , looks like this
$H_{2} S_{(g)} + O_{2 (g)} \to S O_{2 (g)} + H_{2} O_{(g)}$
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 sulphur 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 sulphur dioxide molecule by 2, and the water molecule by 2
$H_{2} S_{(g)} + 3 O_{2 (g)} \to 2 S O_{2 (g)} + 2 H_{2} O_{(g)}$
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
$2 H_{2} S_{(g)} + 3 O_{2 (g)} \to 2 S O_{2 (g)} + 2 H_{2} O_{(g)}$
The hydrogen atoms are balanced, since you ahve 4 on the lefthand side and 48 on the rightand side of the equation.
All the atoms are balanced, so this is how the balanced chemical equation looks like
$2 H_{2} S_{(g)} + 3 O_{2 (g)} \to 2 S O_{2 (g)} + 2 H_{2} O_{(g)}$

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