Question

How do you balance $\mathbf{N}{{\mathbf{a}}_{\mathbf{2}}}\mathbf{S}{{\mathbf{O}}_{\mathbf{4}}}+\mathbf{Ca(NO}{{ }_{\mathbf{3}}}{{\mathbf{)}}_{\mathbf{2}}}\to \mathbf{CaS}{{\mathbf{O}}_{\mathbf{4}}}+\mathbf{NaN}{{\mathbf{O}}_{\mathbf{3}}}$?

Answer 1

Hint: Balancing chemical equations involves the addition of stoichiometric coefficients to the reactants and products. This is important because a chemical equation must obey the law of conservation of mass and the law of constant proportions, i.e., the same number of atoms of each element must exist on the reactant side and the product side of the equation there are two methods of balancing one is traditional method one is algebraic method.

Complete step by step answer:
$\text{N}{{\text{a}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}+\text{Ca(NO}{{ }_{\text{3}}}{{\text{)}}_{\text{2}}}\to \text{CaS}{{\text{O}}_{\text{4}}}+\text{NaN}{{\text{O}}_{\text{3}}}$
By looking at the above reaction we can say that it is a precipitation reaction where insoluble precipitate form as a product.
When balancing a reaction look at the pons, rather than atoms.
We know that
$\text{N}{{\text{a}}_{2}}\text{S}{{\text{O}}_{\text{4}}}\to 2\text{N}{{\text{a}}^{+}}+\text{SO}_{4}^{2-}$
$\text{Ca(N}{{\text{O}}_{3}}{{\text{)}}_{2}}\to \text{C}{{\text{a}}^{2+}}+2\text{NO}_{3}^{-}$
They react and give $\text{CaS}{{\text{O}}_{4}}$ as precipitate and $\text{NaN}{{\text{O}}_{3}}$ as ionic compound.
$\text{NaN}{{\text{O}}_{3}}-\text{N}{{\text{a}}^{+}}+\text{NO}_{3}^{-}$
Now for balancing ions on the reactant side must be equal to the product side.
Here we can see that one calcium ion and one sulphate ion group form $\text{CaS}{{\text{O}}_{4}}$. So no need to balance this.
On other hand $2$ sodium ion on the reactant side and $1$ sodium ion on the product side.
Likewise 2 nitrate ions on the reactant side and $1$ nitrate ions on the product side.
We need to multiply sodium Nitrate by 2 to balance these ions.
So, the complete balanced equation will be
$\text{N}{{\text{a}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}+\text{Ca(NO}{{
}_{\text{3}}}{{\text{)}}_{\text{2}}}\to \text{CaS}{{\text{O}}_{\text{4}}}+2\text{NaN}{{\text{O}}_{\text{3}}}$
Balanced.

Note: In precipitation reactions two soluble ionic compounds to give insoluble precipitates.
In balancing addition of stoichiometric coefficients—describes the total number of molecules of a chemical species that participate in molecules of a chemical species that participate in a chemical reaction. It provides ratios between the reacting species and the product formed in the reaction.