Question

Write the balanced chemical equation for the following.

Concentrated Nitric acid is added to Potassium iodide.

Answer 1

Part 1: Balanced chemical equation:

1. The balanced chemical equation contains the same number of atom on both sides of the reaction.
2. The chemical equation is represented symbolically with the help of the chemical formulae of respective compounds.

Part 2: Steps for balancing

In order to balance this equation, we must have to follow the steps given below.

Step 1: Writing the reaction

1. When Potassium iodide reacts with Nitric acid, it forms Nitrogen dioxide, Potassium nitrate, Water, and Iodine.
2. The unbalanced equation is as follows:

$\underset{(\mathrm{Potassium}\mathrm{iodide})}{\mathrm{KI}\left(\mathrm{s}\right)}+\underset{(\mathrm{Nitric}\mathrm{acid})}{{\mathrm{HNO}}_3(\mathrm{aq}})\to \underset{\left(\mathrm{Iodine}\right)}{{\mathrm{I}}_2\left(\mathrm{g}\right)}+\underset{(\mathrm{Nitrogen}\mathrm{dioxide})}{{\mathrm{NO}}_2\left(\mathrm{g}\right)}+\underset{(\mathrm{Potassium}\mathrm{nitrate})}{{\mathrm{KNO}}_3\left(\mathrm{s}\right)}+{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$

Step 2: Calculating the number of atoms on both the sides of the reaction

1. Now, we first need to identify the number of atoms present in it on both sides i.e. reactant and the product.
2. On the reactant side: 1 atom of Potassium and 1 atom of Iodine is present in Potassium iodide.
3. Similarly, 1 atom of Hydrogen, 1 atom of Nitrogen, and 3 atoms of oxygen are present in Nitric acid.
4. On the product side: 2 atoms of Iodine, 2 atoms of Nitrogen, and 2 atoms of oxygen are present in Nitrogen dioxide.
5. Similarly, 1 atom of Potassium, 1 atom of Nitrogen, and 3 atoms of oxygen are present in Potassium nitrate.
6. Also, 2 atoms of Hydrogen and 1 atoms of oxygen are present in water molecules.

Step 3: Balancing the reactant molecules: by multiplying 2 with Potassium iodide.

1. The chemical reaction can be written as follows:

$\underset{(\mathrm{Potassium}\mathrm{iodide})}{2\mathrm{KI}\left(\mathrm{s}\right)}+\underset{(\mathrm{Nitric}\mathrm{acid})}{{\mathrm{HNO}}_3(\mathrm{aq}})\to \underset{\left(\mathrm{Iodine}\right)}{{\mathrm{I}}_2\left(\mathrm{g}\right)}+\underset{(\mathrm{Nitrogen}\mathrm{dioxide})}{{\mathrm{NO}}_2\left(\mathrm{g}\right)}+\underset{(\mathrm{Potassium}\mathrm{nitrate})}{{\mathrm{KNO}}_3\left(\mathrm{s}\right)}+{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$

Step 4: Balancing the reactant molecules: by multiplying 4 with Nitric acid.

1. The chemical reaction can be written as follows:

$\underset{(\mathrm{Potassium}\mathrm{iodide})}{2\mathrm{KI}\left(\mathrm{s}\right)}+4\underset{(\mathrm{Nitric}\mathrm{acid})}{{\mathrm{HNO}}_3(\mathrm{aq}})\to \underset{\left(\mathrm{Iodine}\right)}{{\mathrm{I}}_2\left(\mathrm{g}\right)}+\underset{(\mathrm{Nitrogen}\mathrm{dioxide})}{{\mathrm{NO}}_2\left(\mathrm{g}\right)}+\underset{(\mathrm{Potassium}\mathrm{nitrate})}{{\mathrm{KNO}}_3\left(\mathrm{s}\right)}+{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$

Step 5: Balancing the product molecules: by multiplying 2 with Nitrogen dioxide.

1. The chemical reaction can be written as follows:

$\underset{(\mathrm{Potassium}\mathrm{iodide})}{2\mathrm{KI}\left(\mathrm{s}\right)}+4\underset{(\mathrm{Nitric}\mathrm{acid})}{{\mathrm{HNO}}_3(\mathrm{aq}})\to \underset{\left(\mathrm{Iodine}\right)}{{\mathrm{I}}_2\left(\mathrm{g}\right)}+\underset{(\mathrm{Nitrogen}\mathrm{dioxide})}{2{\mathrm{NO}}_2\left(\mathrm{g}\right)}+\underset{(\mathrm{Potassium}\mathrm{nitrate})}{{\mathrm{KNO}}_3\left(\mathrm{s}\right)}+{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$

Step 6: Balancing the product molecules: by multiplying 2 with Potassium nitrate.

1. The chemical reaction can be written as follows:

$\underset{(\mathrm{Potassium}\mathrm{iodide})}{2\mathrm{KI}\left(\mathrm{s}\right)}+4\underset{(\mathrm{Nitric}\mathrm{acid})}{{\mathrm{HNO}}_3(\mathrm{aq}})\to \underset{\left(\mathrm{Iodine}\right)}{{\mathrm{I}}_2\left(\mathrm{g}\right)}+\underset{(\mathrm{Nitrogen}\mathrm{dioxide})}{2{\mathrm{NO}}_2\left(\mathrm{g}\right)}+\underset{(\mathrm{Potassium}\mathrm{nitrate})}{2{\mathrm{KNO}}_3\left(\mathrm{s}\right)}+{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$

Step 7: Balancing the product molecules: by multiplying 2 with the Water molecule

1. The chemical reaction can be written as follows:

$\underset{(\mathrm{Potassium}\mathrm{iodide})}{2\mathrm{KI}\left(\mathrm{s}\right)}+4\underset{(\mathrm{Nitric}\mathrm{acid})}{{\mathrm{HNO}}_3(\mathrm{aq}})\to \underset{\left(\mathrm{Iodine}\right)}{{\mathrm{I}}_2\left(\mathrm{g}\right)}+\underset{(\mathrm{Nitrogen}\mathrm{dioxide})}{2{\mathrm{NO}}_2\left(\mathrm{g}\right)}+\underset{(\mathrm{Potassium}\mathrm{nitrate})}{2{\mathrm{KNO}}_3\left(\mathrm{s}\right)}+2{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$

Step 8: Calculate the number of atoms of the balanced equation.

1. On the reactant side: 2 atom of Potassium and 2 atom of Iodine is present in Potassium iodide.
2. Similarly, 4 atoms of Hydrogen, 4 atoms of Nitrogen, and 12 atoms of oxygen are present in Nitric acid.
3. On the product side: 2 atoms of Iodine, 2 atoms of Nitrogen, and 4 atoms of oxygen are present in Nitrogen dioxide.
4. Similarly, 2 atoms of Potassium, 2 atoms of Nitrogen, and 6 atoms of oxygen are present in Potassium nitrate.
5. Also, 4 atoms of Hydrogen and 2 atoms of oxygen are present in water molecules.

Now, the reaction is balanced and the number of atoms is equal on both sides of the reaction.

Thus, the balanced chemical equation is as follows:
$\underset{(\mathrm{Potassium}\mathrm{iodide})}{2\mathrm{KI}\left(\mathrm{s}\right)}+4\underset{(\mathrm{Nitric}\mathrm{acid})}{{\mathrm{HNO}}_3(\mathrm{aq}})\to \underset{\left(\mathrm{Iodine}\right)}{{\mathrm{I}}_2\left(\mathrm{g}\right)}+\underset{(\mathrm{Nitrogen}\mathrm{dioxide})}{2{\mathrm{NO}}_2\left(\mathrm{g}\right)}+\underset{(\mathrm{Potassium}\mathrm{nitrate})}{2{\mathrm{KNO}}_3\left(\mathrm{s}\right)}+2{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$