(iii) `H_(2)SO_(4(aq)) + NaOH_((aq)) to Na_(2)SO_(4(aq)) + H_(2)O_((I))`

Explanation

Rewrite the given equation as it is
` H_(2)SO_(4(aq)) + NaOH_((aq)) to Na_(2)SO_(4(aq)) +H_(2)O_((l))`
Step 2 : Write the number of atoms of each element in the unbalanced equation on both sides of the equation .

The number of oxygen atoms involved in different compounds on both sides (reactants and products) are equal. Therefore, balance the number of atoms of the second element, sodium.
Step 3 : To balance the number of sodium atoms :

To equalise the number of sodium atoms, we use 2 as the factor of NaOH in the reactants. Now, the partly balanced equation becomes as follows :
` H_(2)SO_(4) + 2NaOH to Na_(2)SO_(4) +H_(2)O`
Step 4 : Now, balance the number of hydrogen atoms :

To equalise the number of hydrogen atoms, we use 2 as the factor of ` H_(2)O` in the products. The equation then becomes
`H_(2)SO_(4) + 2NaOH to Na_(2)SO_(4) + 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
` H_(2)SO_(4(aq)) + 2NaOH_((aq)) to Na_(2)SO_(4) + 2H_(2)O_((l))`