Question

For the following reaction, identify the products formed and balance the equation:

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

Answer 1

The reaction can be solved in two steps:

1. Complete the reaction by writing the products.
2. Balance the equation.

1.Complete the reaction by writing the products:

$$\begin{gathered} \mathrm{S}+{\mathrm{HNO}}_3\to {\mathrm{H}}_2{\mathrm{SO}}_4+{\mathrm{NO}}_2+{\mathrm{H}}_2\mathrm{O} \\ \left(\mathrm{Sulphur}\right)(\mathrm{Nitric}\mathrm{acid})(\mathrm{Sulphuric}\mathrm{acid})(\mathrm{Nitrogen}\mathrm{dioxide})\left(\mathrm{Water}\right) \end{gathered}$$

2. Balancing of the reaction by hit and trial method:

A chemical equation is said to be balanced when atoms of every element in the chemical equation are equal in number on both (reactant and product) sides.

To balance the chemical equation, begin by multiplying the element or compound which has less number of atoms with a number that would make the number equal on both sides.

$$\begin{gathered} \mathrm{S}+{\mathrm{HNO}}_3\to {\mathrm{H}}_2{\mathrm{SO}}_4+{\mathrm{NO}}_2+{\mathrm{H}}_2\mathrm{O} \\ \left(\mathrm{Sulphur}\right)(\mathrm{Nitric}\mathrm{acid})(\mathrm{Sulphuric}\mathrm{acid})(\mathrm{Nitrogen}\mathrm{dioxide})\left(\mathrm{Water}\right) \end{gathered}$$

Left hand side (LHS) is the reactant side and Right hand side (RHS) is the product side.

Type of atoms	Number of atoms on L.H.S.	Number of atoms on R.H.S.
Sulphur (S)	1	1
Oxygen (O)	3	7
Hydrogen (H)	1	4
Nitrogen (N)	1	1

Balance the equation by multiplying Nitric acid (HNO₃) on LHS with 6 as the number of atoms of Hydrogen and Oxygen are less on LHS.

$$\begin{gathered} \mathrm{S}+6{\mathrm{HNO}}_3\to {\mathrm{H}}_2{\mathrm{SO}}_4+{\mathrm{NO}}_2+{\mathrm{H}}_2\mathrm{O} \\ \left(\mathrm{Sulphur}\right)(\mathrm{Nitric}\mathrm{acid})(\mathrm{Sulphuric}\mathrm{acid})(\mathrm{Nitrogen}\mathrm{dioxide})\left(\mathrm{Water}\right) \end{gathered}$$

Type of atoms	Number of atoms on L.H.S.	Number of atoms on R.H.S.
Sulphur (S)	1	1
Oxygen (O)	18	7
Hydrogen (H)	6	4
Nitrogen (N)	6	1

As the number of atoms is not equal on both sides, multiply Nitrogen dioxide (NO₂) with 6 as now number of Nitrogen are less on RHS.

$$\begin{gathered} \mathrm{S}+6{\mathrm{HNO}}_3\to {\mathrm{H}}_2{\mathrm{SO}}_4+6{\mathrm{NO}}_2+{\mathrm{H}}_2\mathrm{O} \\ \left(\mathrm{Sulphur}\right)(\mathrm{Nitric}\mathrm{acid})(\mathrm{Sulphuric}\mathrm{acid})(\mathrm{Nitrogen}\mathrm{dioxide})\left(\mathrm{Water}\right) \end{gathered}$$

Type of atoms	Number of atoms on L.H.S.	Number of atoms on R.H.S.
Sulphur (S)	1	1
Oxygen (O)	18	17
Hydrogen (H)	6	4
Nitrogen (N)	6	6

As the number of atoms is still not equal on both sides, multiply water (H₂O) with 2.

$$\begin{gathered} \mathrm{S}+6{\mathrm{HNO}}_3\to {\mathrm{H}}_2{\mathrm{SO}}_4+6{\mathrm{NO}}_2+2{\mathrm{H}}_2\mathrm{O} \\ \left(\mathrm{Sulphur}\right)(\mathrm{Nitric}\mathrm{acid})(\mathrm{Sulphuric}\mathrm{acid})(\mathrm{Nitrogen}\mathrm{dioxide})\left(\mathrm{Water}\right) \end{gathered}$$

Type of atoms	Number of atoms on L.H.S.	Number of atoms on R.H.S.
Sulphur (S)	1	1
Oxygen (O)	18	18
Hydrogen (H)	6
Nitrogen (N)	6	1

As the number of atoms of all elements is now balanced, the equation is now balanced.

So, the overall complete and balanced chemical equation is

$$\begin{gathered} \mathrm{S}\left(\mathrm{aq}\right)+6{\mathrm{HNO}}_3\left(\mathrm{aq}\right)\to {\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)+6{\mathrm{NO}}_2\left(\mathrm{g}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right) \\ \left(\mathrm{Sulphur}\right)(\mathrm{Nitric}\mathrm{acid})(\mathrm{Sulphuric}\mathrm{acid})(\mathrm{Nitrogen}\mathrm{dioxide})\left(\mathrm{Water}\right) \end{gathered}$$