Write the balanced equations for the following reactions :
`CH_(4)+O_(2)rarrCO_(2)+H_(2)O`

Explanation

Step 1 : Rewrite the given equation as it is
`CH_(4)+O_(2)rarrCO_(2)+H_(2)O`
Step 2 : Write the number of atoms of each element in the unbalanced equation on both sides of the equation .

Step 3 : To balance the number of oxygen atoms

To equalise the number of oxygen atoms , we use 2 gas the coefficient of `O_(2)` in the reactants and 2 as the cofficient of `H_(2)O` in the product.
Now, the partly balanced equation becomes
`CH_(4)+2O_(2)rarrCO_(2)+2H_(2)O`
Now , count the atoms of each element on both sides of the equation .The number of atoms on both sides are equal . Hence , the balanced equation is ,
`CH_(4)+2O_(2)rarrCO_(2)+2H_(2)O`
Now , indicate the physical states of the reactant and the products .
`CH_(4_((g)))+2O_(2_((g)))rarrCO_(2_((g)))+2H_(2)O_((1))`