Write the balanced equations for the following reactions : (viii) `KCN+H_(2)SO_(4)rarrK_(2)SO_(4)+HCN`

Explanation

Step 1 : Rewrite the given equation as it is
`KCN+H_(2)SO_(4)rarrK_(2)SO_(4)+HCN`
Step 2 : Write the number of atoms of each element or group in the unbalanced equation on both sides of the equation.

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

To equalise the number of potassium atoms, we use 2 as the cofficient of KCN in the reactants .
Now, the partly balanced equation becomes
`2KCN+H_(2)SO_(4)rarrK_(2)SO_(4)+2HCN`
Now, count the atoms of each element on both sides of the wquation. The nuber of atoms on both sides are equal . Hence ,the balanced equation is
`2KCN+H_(2)SO_(4)rarrK_(2)SO_(4)+2HCN`
Now indicate the physical states of the reactants and the products.
`2KCN_((aq))+H_(2)SO_(4_((aq)))rarrK_(2)SO_(4_((aq)))+2HCN_((g))`