# How would you balance Na2CO3(s) + HCl(aq) = NaCl(aq) + H2O(l) + CO2(g)?

Nov 3, 2015

$N {a}_{2} C {O}_{3} \left(s\right) + \textcolor{red}{2} H C l \left(a q\right) \to \textcolor{red}{2} N a C l \left(a q\right) + {H}_{2} O \left(l\right) + C {O}_{2} \left(g\right)$

#### Explanation:

$N {a}_{2} C {O}_{3} \left(s\right) + \textcolor{red}{2} H C l \left(a q\right) \to \textcolor{red}{2} N a C l \left(a q\right) + {H}_{2} O \left(l\right) + C {O}_{2} \left(g\right)$

The reasoning that you should follow in order to balance this reaction easily is:

1- Look at the $N a$. You have $2$ in $N {a}_{2} C {O}_{3}$ and $1$ in $N a C l$ then you should multiply $N a C l$ by $\textcolor{red}{2}$.

2- When multiplying $N a C l$ by 2, we will have to $C l$ in the products side, therefore, we should multiply $H C l$ by $\textcolor{red}{2}$.

3- Look at the carbon atom $C$ there is one atom at each side. No need for any action.

4- Look at the hydrogen atom $H$ there is 2 atoms at each side. No need for any action.

5- Look at the oxygen atom $O$ there is 3 atoms in $N {a}_{2} C {O}_{3}$ and there is 1 atom in ${H}_{2} O$ and 2 atoms in $C {O}_{2}$ which makes it a total of 3 atoms. No need for any action.