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# Balance the following chemical equations:A.${\text{HN}}{{\text{O}}_{\text{3}}}{\text{ + Ca}}{\left( {{\text{OH}}} \right)_{\text{2}}} \to {\text{Ca}}{\left( {{\text{N}}{{\text{O}}_{\text{3}}}} \right)_{\text{2}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}}$B.${\text{NaOH + }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} \to {\text{N}}{{\text{a}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{0}}$C.${\text{NaCl + AgN}}{{\text{O}}_{\text{3}}} \to {\text{AgCl + NaN}}{{\text{O}}_{\text{3}}}$D.${\text{BaC}}{{\text{l}}_{\text{2}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} \to {\text{BaS}}{{\text{O}}_{\text{4}}}{\text{ + HCl}}$

Last updated date: 29th Mar 2023
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Hint: We also remember that in chemistry, equations play a major role. In the reaction three things are main. There are reactant, product and reaction conditions. The reactants are always in the left side of the equation. If more than one reactant in the equation means by using plus sign. The products are always on the right side of the reaction. Something here also follows. More than one product by using plus signs. In between reactant products are used arrows to find the reaction direction in the equation.

In chemistry, balances the chemical equation means,
We adjust the number moles of reactant and product used to attain equal number of the same atom in reactant and product in the given chemical reaction.
${\text{HN}}{{\text{O}}_{\text{3}}}{\text{ + Ca}}{\left( {{\text{OH}}} \right)_{\text{2}}} \to {\text{Ca}}{\left( {{\text{N}}{{\text{O}}_{\text{3}}}} \right)_{\text{2}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}}$
The given chemical reaction is,
${\text{HN}}{{\text{O}}_{\text{3}}}{\text{ + Ca}}{\left( {{\text{OH}}} \right)_{\text{2}}} \to {\text{Ca}}{\left( {{\text{N}}{{\text{O}}_{\text{3}}}} \right)_{\text{2}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}}$
In the above chemical reaction, except calcium ions are in unbalanced count. Hence, we increases the number of moles of ${\text{HN}}{{\text{O}}_{\text{3}}}$ from one to two in reactant side and the number moles of water also increases from one to two in product side of the chemical reaction.
${\text{2HN}}{{\text{O}}_{\text{3}}}{\text{ + Ca}}{\left( {{\text{OH}}} \right)_{\text{2}}} \to {\text{Ca}}{\left( {{\text{N}}{{\text{O}}_{\text{3}}}} \right)_{\text{2}}}{\text{ + 2}}{{\text{H}}_{\text{2}}}{\text{O}}$
Now, in the above chemical reaction all the atoms are present in equal number in both reactant and product side. So, it has become a balanced chemical reaction.
${\text{NaOH + }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} \to {\text{N}}{{\text{a}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}}$
The given chemical reaction is,
${\text{NaOH + }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} \to {\text{N}}{{\text{a}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}}$
In the above chemical reaction is not balanced, because the atom count in the reactant and product side is not the same number. That the reason we increases the number of moles of ${\text{NaOH}}$ from one to four and increase the number of mole of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ from one to two in the reactant side. In product side, we increase the number of moles of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ from one to two and the number moles of ${{\text{H}}_{\text{2}}}{\text{O}}$ from one to four in the product side of the reaction.
${\text{4NaOH + 2}}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} \to 2{\text{N}}{{\text{a}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{ + 4}}{{\text{H}}_{\text{2}}}{\text{O}}$
Now, all the atoms in the chemical reactions have become equal in reactant and product side of the chemical reaction.
${\text{NaCl + AgN}}{{\text{O}}_{\text{3}}} \to {\text{AgCl + NaN}}{{\text{O}}_{\text{3}}}$
The given chemical reaction is,
${\text{NaCl + AgN}}{{\text{O}}_{\text{3}}} \to {\text{AgCl + NaN}}{{\text{O}}_{\text{3}}}$
The above chemical reaction is considered a balanced chemical reaction. Because the number of silver, sodium, nitrate ion and chlorine atoms are present in the same number in the reactant and product side of the chemical reaction.
${\text{BaC}}{{\text{l}}_{\text{2}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} \to {\text{BaS}}{{\text{O}}_{\text{4}}}{\text{ + HCl}}$
The given chemical reaction is,
${\text{BaC}}{{\text{l}}_{\text{2}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} \to {\text{BaS}}{{\text{O}}_{\text{4}}}{\text{ + HCl}}$
In this chemical reaction, barium and sulphate ions in both sides are equal. But the hydrogen and chlorine atoms are less counted in the product side. That’s the reason we increase the number moles of ${\text{HCl}}$ in the product side from one to two, it will come to a balanced chemical equation.
${\text{BaC}}{{\text{l}}_{\text{2}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} \to {\text{BaS}}{{\text{O}}_{\text{4}}}{\text{ + 2HCl}}$
Now, barium, sulphate ion, hydrogen and chlorine are present in the reactant and product side are equal. So, it is a balanced chemical equation.

Note:
We have to know that in balanced chemical reaction wise, we predict the side product of the reaction. In some cases we are not able to attain the balance by changing the mole. In that case we use the number of countable ions in the product and reactant side of the chemical reaction. In chemistry, redox reaction is one of the types of major reaction. In this redox reaction two methods are used to attain the balanced chemical reaction. There are the ion-electron method and the oxidation number method.