Question

Write the balanced chemical equation for the following:

Concentrated Nitric acid is added to the solid Phosphorous.

Answer 1

Part 1: Balanced chemical equation

1. The chemical equation that contains a similar number of molecules on both sides of an equation or we can say that number of molecules of the reactants used in the equation is equal to the number of molecules of products is termed as balanced chemical equation.
2. They are represented symbolically by the use of chemical formulas with their respective coefficients.

Part 2: Steps for balancing

Step 1: Writing the chemical reaction

1. When concentrated Nitric acid is added to the solid Phosphorous we get Phosphoric acid, Nitrogen dioxide, and Water as the products.
2. The unbalanced chemical equation is written as follows:

$\underset{\left(\mathrm{Phosphorous}\right)}{\mathrm{P}\left(\mathrm{s}\right)}+\underset{(\mathrm{Nitric}\mathrm{acid})}{{\mathrm{HNO}}_3\left(aq\right)}\to \underset{\left(\mathrm{Phosphoric}\mathrm{acid}\right)}{{\mathrm{H}}_3{\mathrm{PO}}_4\left(\mathrm{s}\right)}+\underset{(\mathrm{Nitrogen}\mathrm{dioxide})}{{\mathrm{NO}}_2\left(\mathrm{g}\right)}+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

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

1. On the reactants side: 1 atom of phosphorous
2. Similarly, 1 atom of Hydrogen, 1 atom of Nitrogen, and 3 atoms of Oxygen are present in Nitric acid (${\mathbf{HNO}}_{\mathbf{3}}$).
3. On the products side: 3 atoms of Hydrogen, 1 atom of Phosphorous, and 4 atoms of Oxygen are present in Phosphoric acid $\left(H_3{\mathrm{PO}}_4\right)$.
4. Similarly, 1 atom of Nitrogen and 2 atoms of Oxygen are present in Nitrogen dioxide $\left({\mathrm{NO}}_2\right)$ and 2 atoms of Hydrogen and 1 atom of Oxygen are present in the Water molecule.

Step 3: To balance the chemical equation, we multiply "5" to Nitric acid on the reactant side.

1. The chemical reaction is as follows:

$\underset{\left(\mathrm{Phosphorous}\right)}{\mathrm{P}\left(\mathrm{s}\right)}+\underset{(\mathrm{Nitric}\mathrm{acid})}{5{\mathrm{HNO}}_3\left(\mathrm{l}\right)}\to \underset{\left(\mathrm{Phosphoric}\mathrm{acid}\right)}{{\mathrm{H}}_3{\mathrm{PO}}_4\left(\mathrm{s}\right)}+\underset{(\mathrm{Nitrogen}\mathrm{dioxide})}{{\mathrm{NO}}_2\left(\mathrm{g}\right)}+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

Step 4: To balance the chemical equation, we multiply "5" to Nitrogen dioxide on the product side.

1. The chemical reaction is as follows:
   $\underset{\left(\mathrm{Phosphorous}\right)}{\mathrm{P}\left(\mathrm{s}\right)}+\underset{(\mathrm{Nitric}\mathrm{acid})}{5{\mathrm{HNO}}_3\left(\mathrm{l}\right)}\to \underset{\left(\mathrm{Phosphoric}\mathrm{acid}\right)}{{\mathrm{H}}_3{\mathrm{PO}}_4\left(\mathrm{s}\right)}+\underset{(\mathrm{Nitrogen}\mathrm{dioxide})}{5{\mathrm{NO}}_2\left(\mathrm{g}\right)}+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

Step 5: Calculating the number of molecules on both sides of the reaction.

1. On the reactants side: 1 atom of Phosphorous
2. Similarly, 5 atom of Hydrogen, 5 atom of Nitrogen, and 15 atoms of Oxygen are present in Nitric acid.
3. On the products side: 3 atoms of Hydrogen, 1 atom of Phosphorous, and 4 atoms of Oxygen are present in Phosphoric acid.
4. Similarly, 5 atom of Nitrogen and 10 atoms of Oxygen are present in Nitrogen oxide and 2 atoms of Hydrogen and 1 atom of Oxygen are present in the Water molecule.

Now, the chemical reaction is balanced i.e. number of molecules of reactants is equal to the number of molecules of products.

$\underset{\left(\mathrm{Phosphorous}\right)}{\mathrm{P}\left(\mathrm{s}\right)}+\underset{(\mathrm{Nitric}\mathrm{acid})}{5{\mathrm{HNO}}_3\left(\mathrm{l}\right)}\to \underset{\left(\mathrm{Phosphoric}\mathrm{acid}\right)}{{\mathrm{H}}_3{\mathrm{PO}}_4\left(\mathrm{s}\right)}+\underset{(\mathrm{Nitrogen}\mathrm{dioxide})}{5{\mathrm{NO}}_2\left(\mathrm{l}\right)}+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$