Question

28. Complete and balance the following equation:

• ${\mathrm{K}}_2{\mathrm{CO}}_3+\hspace{0.17em}\mathrm{HCl}\to$

Answer 1

Step 1: Write the skeletal equation for the given reaction

Potassium carbonate (K₂CO₃) reacts with Hydrochloric acid (HCl) and forms Potassium chloride (KCl), Water (H₂O), and Carbon dioxide gas (CO₂). The skeletal chemical equation is as follows:

$\underset{\mathrm{Potassium}\mathrm{carbonate}}{{\mathrm{K}}_2{\mathrm{CO}}_3\left(\mathrm{s}\right)}+\underset{\mathrm{Hydrochloric}\mathrm{acid}}{\mathrm{HCl}\left(\mathrm{aq}\right)}\to \underset{\mathrm{Potassium}\mathrm{chloride}}{\mathrm{KCl}\left(\mathrm{aq}\right)}+\hspace{0.17em}\underset{\mathrm{Water}}{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}\underset{\mathrm{Carbon}\mathrm{dioxide}}{+\hspace{0.17em}{\mathrm{CO}}_2\left(\mathrm{g}\right)}$

Step 2: Count the number of atoms of each element on both sides of the equation

For balancing the equation, both reactants and products should be balanced on both sides. The number of atoms of each element present on both sides of the equation is listed below in the table:

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

Step 3: Balance the number of H atoms on the reactant side

• Since it is clear that the number of atoms on reactants and product sides is not equal.
• Thus, for balancing H-atoms, multiply HCl on the product side by 2.

$$\begin{gathered} \underset{\mathrm{Potassium}\mathrm{carbonate}}{{\mathrm{K}}_2{\mathrm{CO}}_3}+2\times \left(\underset{\mathrm{Hydrochloric}\mathrm{acid}}{\hspace{0.17em}\mathrm{HCl}}\right)\to \underset{\mathrm{Potassium}\mathrm{chloride}}{\mathrm{KCl}}+\hspace{0.17em}\underset{\mathrm{Water}}{{\mathrm{H}}_2\mathrm{O}}\underset{\mathrm{Carbon}\mathrm{dioxide}}{+\hspace{0.17em}{\mathrm{CO}}_2\uparrow } \\ \underset{\mathrm{Potassium}\mathrm{carbonate}}{{\mathrm{K}}_2{\mathrm{CO}}_3}+\underset{\mathrm{Hydrochloric}\mathrm{acid}}{\hspace{0.17em}2\mathrm{HCl}}\to \underset{\mathrm{Potassium}\mathrm{chloride}}{\mathrm{KCl}}+\hspace{0.17em}\underset{\mathrm{Water}}{{\mathrm{H}}_2\mathrm{O}}\underset{\mathrm{Carbon}\mathrm{dioxide}}{+\hspace{0.17em}{\mathrm{CO}}_2\uparrow } \end{gathered}$$

Step 4: Balance the number of K and Cl-atoms on the product side

• From the above, it is clear that atoms of potassium and chlorine are not balanced.
• For balancing K and Cl-atoms, multiply KCl on the product side by 2.

$$\begin{gathered} \underset{\mathrm{Potassium}\mathrm{carbonate}}{{\mathrm{K}}_2{\mathrm{CO}}_3}+\underset{\mathrm{Hydrochloric}\mathrm{acid}}{\hspace{0.17em}\mathrm{HCl}}\to 2\times \left(\underset{\mathrm{Potassium}\mathrm{chloride}}{\mathrm{KCl}}\right)+\hspace{0.17em}\underset{\mathrm{Water}}{{\mathrm{H}}_2\mathrm{O}}\underset{\mathrm{Carbon}\mathrm{dioxide}}{+\hspace{0.17em}{\mathrm{CO}}_2\uparrow } \\ \underset{\mathrm{Potassium}\mathrm{carbonate}}{{\mathrm{K}}_2{\mathrm{CO}}_3}+\underset{\mathrm{Hydrochloric}\mathrm{acid}}{\hspace{0.17em}2\mathrm{HCl}}\to \underset{\mathrm{Potassium}\mathrm{chloride}}{2\mathrm{KCl}}+\hspace{0.17em}\underset{\mathrm{Water}}{{\mathrm{H}}_2\mathrm{O}}\underset{\mathrm{Carbon}\mathrm{dioxide}}{+\hspace{0.17em}{\mathrm{CO}}_2\uparrow } \end{gathered}$$

Since the number of atoms is equal on both sides thus, the above equation is said to be balanced.

Therefore, the balanced chemical equation for the given reaction is as follows:

$\underset{\mathrm{Potassium}\mathrm{carbonate}}{{\mathrm{K}}_2{\mathrm{CO}}_3\left(\mathrm{s}\right)}+\underset{\mathrm{Hydrochloric}\mathrm{acid}}{2\mathrm{HCl}\left(\mathrm{aq}\right)}\to \underset{\mathrm{Potassium}\mathrm{chloride}}{2\mathrm{KCl}\left(\mathrm{aq}\right)}+\hspace{0.17em}\underset{\mathrm{Water}}{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}\underset{\mathrm{Carbon}\mathrm{dioxide}}{+\hspace{0.17em}{\mathrm{CO}}_2\left(\mathrm{g}\right)}$