Question

Complete the balance equation given below:

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

Answer 1

Step1: Reacting to the given molecules:

1. When Sulphur reacts with Nitric acid it oxidizes Sulphur to form Sulphur dioxide which further proceeds to form Sulphur trioxide.
2. In the end we get Sulphuric acid as the final product.
3. The unbalanced chemical equation can be written as:$\underset{\left(\mathrm{Sulphur}\right)}{\mathrm{S}\left(\mathrm{s}\right)}+\underset{(\mathrm{Nitric}\mathrm{acid})}{{\mathrm{HNO}}_3\left(\mathrm{l}\right)}\to \underset{(\mathrm{Sulphuric}\mathrm{acid})}{{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{l}\right)}+{\mathrm{H}}_2\mathrm{O}+\underset{(\mathrm{Nitrogen}\mathrm{dioxide})}{{\mathrm{NO}}_2\left(\mathrm{g}\right)}$

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

Step 2: Balancing the chemical equation

1. The total number of atoms of each element on the reactant side and the product side must be compared.
2. On the reactant side: 1 Hydrogen, 1 Nitrogen and 3 Oxygen atoms are present in Nitric acid
3. On the product side: 2 Hydrogen, 1 Sulphur and 4 Oxygen is present in Sulphuric acid.
4. Similarly; Nitrogen dioxide contains 1 Nitrogen 2 oxygen and 2 Hydrogen and 1 Oxygen atoms are present in the water molecule.

Step 2: Balancing the reactant molecules: Nitric acid is multiplied by 6.

1. The equation becomes: $\underset{\left(\mathrm{Sulphur}\right)}{\mathrm{S}\left(\mathrm{s}\right)}+\underset{(\mathrm{Nitric}\mathrm{acid})}{6{\mathrm{HNO}}_3\left(\mathrm{l}\right)}\to \underset{(\mathrm{Sulphuric}\mathrm{acid})}{{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{l}\right)}+{\mathrm{H}}_2\mathrm{O}+\underset{(\mathrm{Nitrogen}\mathrm{dioxide})}{{\mathrm{NO}}_2\left(\mathrm{g}\right)}$

Step 3: Balancing the product molecules: The water molecule is multiplied by 2.

1. The equation becomes: $\underset{\left(\mathrm{Sulphur}\right)}{\mathrm{S}\left(\mathrm{s}\right)}+\underset{(\mathrm{Nitric}\mathrm{acid})}{6{\mathrm{HNO}}_3\left(\mathrm{l}\right)}\to \underset{(\mathrm{Sulphuric}\mathrm{acid})}{{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{l}\right)}+2{\mathrm{H}}_2\mathrm{O}+\underset{(\mathrm{Nitrogen}\mathrm{dioxide})}{{\mathrm{NO}}_2\left(\mathrm{g}\right)}$

Step 3: Balancing the product molecules: Nitrogen dioxide is multiplied by 2.

1. The equation becomes: $\underset{\left(\mathrm{Sulphur}\right)}{\mathrm{S}\left(\mathrm{s}\right)}+\underset{(\mathrm{Nitric}\mathrm{acid})}{6{\mathrm{HNO}}_3\left(\mathrm{l}\right)}\to \underset{(\mathrm{Sulphuric}\mathrm{acid})}{{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{l}\right)}+2{\mathrm{H}}_2\mathrm{O}+\underset{(\mathrm{Nitrogen}\mathrm{dioxide})}{2{\mathrm{NO}}_2\left(\mathrm{g}\right)}$

Now, the total number of molecules are equal on both sides of the reaction.

1. On the reactant side: 6 Hydrogen, 6 Nitrogen and 18 Oxygen atoms are present in Nitric acid
2. On the product side: 2 Hydrogen, 1 Sulphur and 4 Oxygen are present in Sulphuric acid.
3. Similarly; Nitrogen dioxide contains 6 Nitrogen 12 oxygen and 4 Hydrogen and 2 Oxygen atoms are present in the water molecule.

Hence we get the chemical equation is as follows:$\underset{\left(\mathrm{Sulphur}\right)}{\mathrm{S}\left(\mathrm{s}\right)}+\underset{(\mathrm{Nitric}\mathrm{acid})}{6{\mathrm{HNO}}_3\left(\mathrm{l}\right)}\to \underset{(\mathrm{Sulphuric}\mathrm{acid})}{{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{l}\right)}+2{\mathrm{H}}_2\mathrm{O}+\underset{(\mathrm{Nitrogen}\mathrm{dioxide})}{6{\mathrm{NO}}_2\left(\mathrm{g}\right)}$