Question

Complete and balance the following reaction:

$\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2+2\mathrm{HCl}(\mathrm{dil}.)\to$

Answer 1

• When dilute Hydrochloric acid is added to Lead Nitrate solution, it gives a white precipitate of Lead chloride and Nitric acid
• The chemical reaction is as follows:
  $$\begin{gathered} \underset{(\mathrm{Lead}\mathrm{nitrate})}{\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2\left(\mathrm{s}\right)}+\underset{(\mathrm{Hydrogen}\mathrm{chloride})}{\mathrm{HCl}\left(\mathrm{aq}\right)}\to \underset{(\mathrm{Lead}\mathrm{chloride})}{{\mathrm{PbCl}}_2\left(\mathrm{s}\right)\downarrow }+\underset{(\mathrm{Nitric}\mathrm{acid}) \\ }{{\mathrm{HNO}}_3\left(\mathrm{aq}\right)} \end{gathered}$$

The obtained reaction is unbalanced in nature and to balance such a reaction we need to calculate the number of molecules.

Step 1: Balancing the equation:

• Calculating the number of molecules on the reactant side: 1 Lead, 2 Nitrogen, and 6 Oxygen atoms are present in Lead nitrate.
• Similarly; 1 Hydrogen and 1 Chlorine are present in Hydrogen chloride.
• Calculating the number of molecules on the product side: 1 Lead and 2 Chlorine atoms are present in the Lead chloride molecule.
• Similarly; 1 Hydrogen, 1 Nitrogen, and 3 Oxygen atoms are present in Nitric acid.

Step 2: Balancing the reactant molecules: Multiplying Hydrogen chloride with 2.

• The chemical reaction is as follows: $$\begin{gathered} \underset{(\mathrm{Lead}\mathrm{nitrate})}{\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2\left(\mathrm{s}\right)}+\underset{(\mathrm{Hydrogen}\mathrm{chloride})}{2\mathrm{HCl}\left(\mathrm{aq}\right)}\to \underset{(\mathrm{Lead}\mathrm{chloride})}{{\mathrm{PbCl}}_2\left(\mathrm{s}\right)\downarrow }+\underset{(\mathrm{Nitric}\mathrm{acid}) \\ }{{\mathrm{HNO}}_3\left(\mathrm{aq}\right)} \end{gathered}$$

Step 3: Balancing the product molecules: Multiplying Nitric acid with 2.

• The chemical reaction is as follows: $$\begin{gathered} \underset{(\mathrm{Lead}\mathrm{nitrate})}{\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2\left(\mathrm{s}\right)}+\underset{(\mathrm{Hydrogen}\mathrm{chloride})}{2\mathrm{HCl}\left(\mathrm{aq}\right)}\to \underset{(\mathrm{Lead}\mathrm{chloride})}{{\mathrm{PbCl}}_2\left(\mathrm{s}\right)\downarrow }+2\underset{(\mathrm{Nitric}\mathrm{acid}) \\ }{{\mathrm{HNO}}_3\left(\mathrm{aq}\right)} \end{gathered}$$

Thus, we get a balanced equation where the total number of molecules on the reactant side is equal to the product side.

• the number of molecules on the reactant side: 1 Lead, 2 Nitrogen, and 6 Oxygen atoms are present in Lead nitrate.
• Similarly; 2 Hydrogen and 2 Chlorine are present in Hydrogen chloride.
• the number of molecules on the reactant side: 1 Lead and 2 Chlorine atoms are present in the Lead chloride molecule.
• Similarly; 2 Hydrogen, 2 Nitrogen, and 6 Oxygen atoms are present in Nitric acid.

Therefore, both the sides of the molecules are balanced and the chemical equation is as follows:

$$\begin{gathered} \underset{(\mathrm{Lead}\mathrm{nitrate})}{\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2\left(\mathrm{s}\right)}+\underset{(\mathrm{Hydrogen}\mathrm{chloride})}{2\mathrm{HCl}\left(\mathrm{aq}\right)}\to \underset{(\mathrm{Lead}\mathrm{chloride})}{{\mathrm{PbCl}}_2\left(\mathrm{s}\right)\downarrow }+2\underset{(\mathrm{Nitric}\mathrm{acid}) \\ }{{\mathrm{HNO}}_3\left(\mathrm{aq}\right)} \end{gathered}$$