Chapter 19.1, Problem 3PP

To determine

To balance: the equation when HCl is slowly added to aqueous ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ forming NaCl, ${\text{H}}_{\text{2}}\text{O}$ ,

Explanation of Solution

Introduction:

In balance equation, reactant and product will have same number of coefficient. Chemical reactions can be explained in a precise way through chemical equations.

The given reaction is $\text{HCl}\left(\text{aq}\right){\text{+Na}}_{\text{2}}{\text{CO}}_{\text{3}}\left(\text{aq}\right)\to \text{NaCl}\left(\text{aq}\right){\text{+H}}_{\text{2}}\text{O}\left(\text{l}\right){\text{+CO}}_{\text{2}}\left(\text{g}\right)$

Here, in this reaction, all atoms are balanced except $\text{Na}$ and $\text{H}$ .

Balance $\text{Na}$ on the product side by multiplying $\text{NaCl}$ with 2.

$\text{HCl}\left(\text{aq}\right){\text{+Na}}_{\text{2}}{\text{CO}}_{\text{3}}\left(\text{aq}\right)\to 2\text{NaCl}\left(\text{aq}\right){\text{+H}}_{\text{2}}\text{O}\left(\text{l}\right){\text{+CO}}_{\text{2}}\left(\text{g}\right)$

So, $\text{Na}$ is balanced on both the reactant and product side.

Now, balance H and $\text{Cl}$ on reactant side by multiplying $\text{HCl}$ with 2,

$\text{2HCl}\left(\text{aq}\right){\text{+Na}}_{\text{2}}{\text{CO}}_{\text{3}}\left(\text{aq}\right)\to 2\text{NaCl}\left(\text{aq}\right){\text{+H}}_{\text{2}}\text{O}\left(\text{l}\right){\text{+CO}}_{\text{2}}\left(\text{g}\right)$

Thus, the balance reactant is,

$\text{2HCl}\left(\text{aq}\right){\text{+Na}}_{\text{2}}{\text{CO}}_{\text{3}}\left(\text{aq}\right)\to 2\text{NaCl}\left(\text{aq}\right){\text{+H}}_{\text{2}}\text{O}\left(\text{l}\right){\text{+CO}}_{\text{2}}\left(\text{g}\right)$

Conclusion:

Hence, the balanced equation is $\text{2HCl}\left(\text{aq}\right){\text{+Na}}_{\text{2}}{\text{CO}}_{\text{3}}\left(\text{aq}\right)\to 2\text{NaCl}\left(\text{aq}\right){\text{+H}}_{\text{2}}\text{O}\left(\text{l}\right){\text{+CO}}_{\text{2}}\left(\text{g}\right)$ .