Balance the following chemical equations:
(a) `HNO_(3)+Ca(OH)_(2)toCa(NO_(3))_(2)+H_(2)O`
(b) `NaOH+H_(2)SO_(4)toNa_(2)SO_(4)+H_(2)O`
(c) `NaCl+AgNO_(3)toAgCl+NaNO_(3)`
(d) `BaCl_(2)+H_(2)SO_(4)toBaSO_(4)+HCl`

Explanation

(a) The symbol equation as gien for the reaction is :
`HNO_(3) + Ca(OH)_(2) to Ca(NO_(3))_(2) + H_(2)O`
The balancing fo the equation is done in the following steps :
Step I. Let us count the number of atoms of all the elements on both sides of the equation.

A simple look at the equation reveals that the number of Ca atoms are equal on both sides.
Step II. In order to equate the number of N atoms, put coefficient 2 before `HNO_(3)` on the reactant side.
`2HNO_(3) + Ca(OH)_(2) to Ca(NO_(3))_(2) + H_(2)O`
Step III. In order to equate the number of H atoms, put coefficient 2 before `H_(2)O` on the product side.
`2HNO_(3) + Ca(OH)_(2) to Ca(NO_(3))_(2) + 2H_(2)O`
Step IV. On inspection the number of O atoms on both sides of the equation is the same i.e., 8. Therefore, the equation is balanced.
(b) The symbol equation as given for the reaction is :
`NaOH + H_(2)SO_(4) to Na_(2)SO_(4) + H_(2)O`
Step I. Let us count the number of atoms of all the elements on both sides of the equation.

A simple look at the equation reveals that the number of O and s atoms are equal on both sides.
Step II. In order to equate the number of Na atoms, put coefficient 2 before `NaOH` on the reactant side.
`2NaOH + H_(2)SO_(4) to Na_(2)SO_(4) + H_(2)O`
Step III.In order to equate the number of H atoms, put coefficient 2 before `H_(2)O` on the product side.
`2NaOH + H_(2)SO_(4) to Na_(2)SO_(4) + 2H_(2)O`
Step IV. On inspection, the number of O atoms on the both sides of the equation is the same i.e., 6. Therefore, the equation is balanced.
(c) The symbol equation as given for the reaction is already balanced.
`NaCl + AgNO_(3) to AgCl + NaNO_(3)`
On inspection, the number of atoms of all the elements are found to be equal on both sides of the equation. It is in the balanced form.
(d) The symbol equation as given for the reaction is :
`BaCl_(2) + H_(2)SO_(4) to BaSO_(4) + HCl`
Step I. Let us count the number of atoms of all the elements on both sides of the equation.

A simple look at the equation reveals that the number of Ba, s and O atoms are equal on both sides.
Step II. In order to equate the number of Cl atoms, put coefficient 2 before HCl on the product side.
`BaCl_(2) + H_(2)SO_(4) to BaSO_(4) + 2HCl`
Step III. On inspection, the number of H atoms on both sides of the equation is the same i.e., 2. Therefore, the equation is balanced.