Balance the following chemical eqution:
`Fe(s)+H_(2)O(g)rarrFe_(3)O_(4)(s)+H_(2)(g)`

Explanation

This is an example of a reaction which occurs in absence of acids and bases nd hence blancing of O atoms cannot be done by addition of `H_(2)O` molecules but has to be done on the basis of gain or loss of eelctrons to balance such reactions following steps are followed
Step 1 Identify atoms whose oxidation numbers undergo a change writing the oxidation number of each atom above its symbol we have `overset(+2)Feoverset(-1)S_(2)+O_(2) to Fe_(2)O_(3)^(-)+S O_(2)` that of O has decreased form 0 to -2 in other owrds both Fe and S of `FeS_(2)` hav been oxidised since Fe and S must maintain therir atomic ratio of 1:2 therefore the change of oxidation number of these two atoms must be considered together
Step 2 Determine the total increase nd decrease in oxidation number indicating the increase nd decrease in oxidation numbners in each case we have

Step 3 Balance the total increase and decrease in oxidation numbers
To balance the total increase in oxidation number of Fe and S and decrease in oxidation number of O multiply Eq (ii) by 4 nad Eq (iii) by 11 and adding we have
Since `overset(2-)O` does not occur independetly so the `22overset(2-)` must be factorized in such a manner that hey become parts of `Fe_(2)O_(3) and SO_(2)` Rearranging Eq (iv) we have
`4FeS_(2)+11 O_(2) to 2Fe_(2)O_(3)+8 SO_(2)`
This represents the required balanced equation
II Half Reactin mehod or ion electron method
This method of balancing redox equation is based upon the principle that electrons lost during oxidation half rection o fnay redox reaction are equal to the eletrons gained during reduction half rection the various steps involved in this method are
Step 1 Write the skeletal equation and indicate the oxidation number (O.N) of all the elements which appear in the skeletal equation above their respective symbols
Step 2 Find out the specied whihc are oxidaised and which are reduced
Step 3 Split the skeletal equation in to two half reactions i.e oxidaion half reaction and reduction half reation
Step 4 Balacne the two half reaction equatin rsepreately by the rules described below
(i) In each half reaction first balace the atoms of ht elements which have undergone a change in oxidation number
(ii) Add electrons to whatever side is necessary to make up the difference in oxation number in each half reaction
(iii) alacne charge by adding `H^(+)` ions if the reaction occurs in the acidic medium and by adding `OH^(-)` ions if the reaction occurs in the basic medium
(iv) Balance oxygen atoms by adding required number of `H_(2)O` molecules to the side deficient in O atoms
(v)In the acidic medium H atoms are balcned by adding `H^(+)` ions to hte side deficient in H atoms however in an equal number `OH^(-)` ions ae included in the opposidet side of the equation Remove the duplication if any
Step 5 The two half reaction are then multiplied by suitable integers so that the totyal number of elctorns gained in one half reaction is equal to the number of electrons lost in the other half reaction The two half reactions are then added up these rules are illustrated by the following expales
step 6 verification To verify whether the eaution thus obtained is balaced or not the total charge on sither side the equation must be equal