Question

Balance the equation $$\begin{gathered} \mathrm{Cu}\left(\mathrm{s}\right)+{\mathrm{HNO}}_3\left(\mathrm{l}\right)\to \mathrm{Cu}{\left({\mathrm{NO}}_3\right)}_2+{\mathrm{H}}_2\mathrm{O}+{\mathrm{NO}}_2 \\ \left(\mathrm{copper}\right)(\mathrm{nitric}\mathrm{acid})(\mathrm{copper}\mathrm{nitrate})\left(\mathrm{water}\right)(\mathrm{nitrogen}\mathrm{dioxide}) \end{gathered}$$

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:

$$\begin{gathered} \mathrm{Cu}\left(\mathrm{s}\right)+{\mathrm{HNO}}_3\left(\mathrm{l}\right)\to \mathrm{Cu}{\left({\mathrm{NO}}_3\right)}_2+{\mathrm{H}}_2\mathrm{O}+{\mathrm{NO}}_2 \\ \left(\mathrm{copper}\right)(\mathrm{nitric}\mathrm{acid})(\mathrm{copper}\mathrm{nitrate})\left(\mathrm{water}\right)(\mathrm{nitrogen}\mathrm{dioxide}) \end{gathered}$$

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

A balanced equation is a chemical reaction equation in which the overall charge and the number of atoms for each element in the reaction are the same for both the reactants and the products.

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.

$$\begin{gathered} \mathrm{Cu}\left(\mathrm{s}\right)+{\mathrm{HNO}}_3\left(\mathrm{l}\right)\to \mathrm{Cu}{\left({\mathrm{NO}}_3\right)}_2+{\mathrm{H}}_2\mathrm{O}+{\mathrm{NO}}_2 \\ \left(\mathrm{copper}\right)(\mathrm{nitric}\mathrm{acid})(\mathrm{copper}\mathrm{nitrate})\left(\mathrm{water}\right)(\mathrm{nitrogen}\mathrm{dioxide}) \end{gathered}$$

Types of atoms	Number of atoms on the reactant side	Number of atoms on the product side
Copper	1	1
Hydrogen	1	2
Oxygen	3	9
Nitrogen	1	3

Balance the equation by multiplying nitric acid by 2 on the reactant side as the number of hydrogen atoms are less on the reactant side.

$$\begin{gathered} \mathrm{Cu}\left(\mathrm{s}\right)+2{\mathrm{HNO}}_3\left(\mathrm{l}\right)\to \mathrm{Cu}{\left({\mathrm{NO}}_3\right)}_2+{\mathrm{H}}_2\mathrm{O}+{\mathrm{NO}}_2 \\ \left(\mathrm{copper}\right)(\mathrm{nitric}\mathrm{acid})(\mathrm{copper}\mathrm{nitrate})\left(\mathrm{water}\right)(\mathrm{nitrogen}\mathrm{dioxide}) \end{gathered}$$

Types of atoms	Number of atoms on the reactant side	Number of atoms on the product side
Copper	1	1
Hydrogen	2	2
Oxygen	6	9
Nitrogen	2	3

As the number of atoms is still unequal on both sides, multiplying nitric acid, water, and nitrogen dioxide by 2.

$$\begin{gathered} \mathrm{Cu}\left(\mathrm{s}\right)+4{\mathrm{HNO}}_3\left(\mathrm{l}\right)\to \mathrm{Cu}{\left({\mathrm{NO}}_3\right)}_2+2{\mathrm{H}}_2\mathrm{O}+2{\mathrm{NO}}_2 \\ \left(\mathrm{copper}\right)(\mathrm{nitric}\mathrm{acid})(\mathrm{copper}\mathrm{nitrate})\left(\mathrm{water}\right)(\mathrm{nitrogen}\mathrm{dioxide}) \end{gathered}$$

Types of atoms	Number of atoms on the reactant side	Number of atoms on the product side
Copper	1	1
Hydrogen	4	4
Oxygen	12	12
Nitrogen	4	4

As the number of all the atoms is equal on both sides, the chemical equation is now balanced.

So, the overall complete and balanced chemical equation is

$$\begin{gathered} \mathrm{Cu}\left(\mathrm{s}\right)+4{\mathrm{HNO}}_3\left(\mathrm{l}\right)\to \mathrm{Cu}{\left({\mathrm{NO}}_3\right)}_2\left(\mathrm{s}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)+2{\mathrm{NO}}_2\left(\mathrm{g}\right) \\ \left(\mathrm{copper}\right)(\mathrm{nitric}\mathrm{acid})(\mathrm{copper}\mathrm{nitrate})\left(\mathrm{water}\right)(\mathrm{nitrogen}\mathrm{dioxide}) \end{gathered}$$