Question

Complete and balance the equation given below:

$\mathrm{P}+{\mathrm{HNO}}_3\to$

Answer 1

1. Nitric acid$\mathbf{\left(}{\mathbf{HNO}}_{\mathbf{3}}\mathbf{\right)}$ acts as an oxidizing agent.
2. The reaction between Phosphorous and Nitric acid$\mathbf{\left(}{\mathbf{HNO}}_{\mathbf{3}}\mathbf{\right)}$ gives Nitrogen dioxide$\left({\mathrm{NO}}_2\right)$, water, and Phosphoric acid.
3. The Phosphoric acid produces in the reaction is a white crystalline solid.
4. The chemical reaction occur as follows:$\underset{\left(\mathrm{Phosphorous}\right)}{\mathrm{P}\left(\mathrm{s}\right)}+\underset{(\mathrm{Nitric}\mathrm{acid})}{{\mathrm{HNO}}_3\left(\mathrm{l}\right)}\to \underset{(\mathrm{Nitrogen}\mathrm{dioxide})}{{\mathrm{NO}}_2\left(\mathrm{g}\right)}+\underset{(\mathrm{Phosphoric}\mathrm{acid})}{{\mathrm{H}}_3{\mathrm{PO}}_4\left(\mathrm{s}\right)}+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

The obtained equation is unbalanced, to balance this equation we first need to calculate the number of molecules on both sides of the reaction.

Step 1: Counting the number of molecules on both sides.

1. On the reactant side: 1 Hydrogen, 1 Nitrogen, and 3 Oxygen atom are present in Nitric acid.
2. On the product side: 3 Hydrogen, 1 Phosphorous 4 Oxygen atoms are present in Phosphoric acid.
3. Similarly, Nitrogen dioxide contains 1 Nitrogen 2 Oxygen atoms, and 2 Hydrogen and 1 Oxygen atoms are present in the water molecule.

Step 2: Balancing the reactant molecules: Nitric acid is multiplied by 5.

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

Step 2: Balancing the reactant molecules: Nitrogen dioxide is multiplied by 5.

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

Now, the equation is balanced i.e. total number of molecules on both sides of the reaction is equal.

1. On the reactant side: 5 Hydrogen, 5 Nitrogen, and 15 Oxygen atom are present in Nitric acid.
2. On the product side: 3 Hydrogen, 1 Phosphorous 4 Oxygen atoms are present in Phosphoric acid.
3. Similarly, Nitrogen dioxide contains 5 Nitrogen, 10 Oxygen atoms, and 2 Hydrogen and 1 Oxygen atoms are present in the water molecule.

Therefore, the balanced equation is as follows:$\underset{\left(\mathrm{Phosphorous}\right)}{\mathrm{P}\left(\mathrm{s}\right)}+5\underset{(\mathrm{Nitric}\mathrm{acid})}{{\mathrm{HNO}}_3\left(\mathrm{l}\right)}\to \underset{(\mathrm{Nitrogen}\mathrm{dioxide})}{5{\mathrm{NO}}_2\left(\mathrm{g}\right)}+\underset{(\mathrm{Phosphoric}\mathrm{acid})}{{\mathrm{H}}_3{\mathrm{PO}}_4\left(\mathrm{s}\right)}+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$