Question

Balance the following equation step wise.
\[S{{O}_{2}}(g)+{{H}_{2}}S(aq.)\to S(s)+{{H}_{2}}O(l)\]

Answer 1

Hint: So, for balancing any chemical reaction we have to balance the number of moles of each atom at the reactant side or product side.

Complete step by step solution:
 For balancing the given reaction 1st off all we will balance the number of atoms or number of moles of sulphur (S) atom:
So, in the right hand side only one sulphur atom is given but in the left hand side two sulphur atoms are there. So we add one sulphur atom in the left side:
\[S{{O}_{2}}(g)+{{H}_{2}}S(aq.)\to 2S(s)+{{H}_{2}}O(I)\]
Now both sides of the reaction have two sulphur atoms.
Next we will balance the oxygen atom, on the left side there are two oxygen atoms but on the right side only one oxygen atom is there. So we will add one more water molecule:
\[S{{O}_{2}}(g)+{{H}_{2}}S(aq.)\to 2S(s)+2{{H}_{2}}O(I)\]
Now oxygen and sulphur are balanced.
Next we have to balance hydrogen atom right side have 4 hydrogen atoms and left side has 2 hydrogen atoms so, we can add one more \[{{H}_{2}}S\], but if add one more \[{{H}_{2}}S\] than we have to add one sulphur atom :
\[S{{O}_{2}}(g)+2{{H}_{2}}S(aq.)\to 3S(s)+2{{H}_{2}}O(I)\]
There is no need to add one more sulphur atom, because we can start balancing from hydrogen then oxygen and then sulphur, so, we do this sequence then no need to repeat the balancing of sulphur again.
\[\;\;\;\;\;\;\;\;\text{Reactant side} \;\;\;\;\;\;\;\;\;\text{Product side}\]
$S\;\;\;\;\;\;\;\;\;\;\;\;\;\;3\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;3$

$O\;\;\;\;\;\;\;\;\;\;\;\;\;\;2\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;2$

$H\;\;\;\;\;\;\;\;\;\;\;\;\;\;4\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;4$

Note: So, after balancing the reaction we have to again check that the number of each atom at reactant side equal to the number of each atom at left side. If you have a single atom which is not bonded with other atoms, we can do its balancing at the end.