Question

Complete and balance the reaction

${\mathrm{Na}}_2{\mathrm{CO}}_3+\mathrm{HCl}\to$

Answer 1

Step 1: When Sodium carbonate $\left({\mathrm{Na}}_2{\mathrm{CO}}_3\right)$reacts with Hydrochloric acid $\left(\mathrm{HCl}\right)$, then it gives Sodium chloride$\left(\mathrm{NaCl}\right)$, water$\left({\mathrm{H}}_2\mathrm{O}\right)$, and carbon dioxide$\left({\mathrm{CO}}_2\right)$ as a gas.

$N{a}_2{\mathrm{CO}}_3+\mathrm{HCl}\to \mathrm{NaCl}+{\mathrm{H}}_2\mathrm{O}+{\mathrm{CO}}_2$

Step 2: Balancing step by step as:

1. Count the number of atoms in both the reactant and product sides.

Reactant	Product
Na - 2	Na - 1
C - 1	C - 1
O - 3	O - 3
H - 1	H - 2
Cl - 1	Cl - 1

Here, Oxygen, Carbon, and Chlorine are balanced but Sodium and Hydrogen are not balanced on both sides.

2: Multiply $\mathrm{NaCl}$ by 2 on the product side to balance Sodium on both sides.

$N{a}_2{\mathrm{CO}}_3+\mathrm{HCl}\to 2\mathrm{NaCl}+{\mathrm{H}}_2\mathrm{O}+{\mathrm{CO}}_2$

3: Count the number of atoms in both the reactant and product sides.

Reactant	Product
Na - 2	Na - 2
C - 1	C - 1
O - 3	O - 3
H - 1	H - 2
Cl - 1	Cl - 2

Here, Sodium, Oxygen, and Carbon are balanced but Hydrogen and Chlorine are not balanced on both sides.

4: Multiply $\mathrm{HCl}$ by 2 on the reactant side to balance Hydrogen and Chlorine.

$N{a}_2{\mathrm{CO}}_3+2\mathrm{HCl}\to 2\mathrm{NaCl}+{\mathrm{H}}_2\mathrm{O}+{\mathrm{CO}}_2$

5: Count the number of atoms in both the reactant and product sides.

Reactant	Product
Na - 2	Na - 2
C - 1	C - 1
O - 3	O - 3
H - 2	H - 2
Cl - 2	Cl - 2

Now, the equation is balanced as all the atoms are balanced on both sides.
Step 3: $N{a}_2{\mathrm{CO}}_3+2\mathrm{HCl}\to 2\mathrm{NaCl}+{\mathrm{H}}_2\mathrm{O}+{\mathrm{CO}}_2$