Question

Balance the following chemical reaction by identifying the products:

${\mathrm{Fe}}_3{\mathrm{O}}_4+\mathrm{HCl}\to {\mathrm{FeCl}}_2+{\mathrm{FeCl}}_3+\mathrm{\_\_\_\_\_}$

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:

$$\begin{gathered} {\mathrm{Fe}}_3{\mathrm{O}}_4+\mathrm{HCl}\to {\mathrm{FeCl}}_2+{\mathrm{FeCl}}_3+{\mathrm{H}}_2\mathrm{O} \\ (\mathrm{Ferroso}-\mathrm{ferric}\mathrm{oxide})(\mathrm{Hydrochloric}\mathrm{acid})(\mathrm{Ferrous}\mathrm{chloride})(\mathrm{Ferric}\mathrm{chloride})\left(\mathrm{Water}\right) \end{gathered}$$

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.

$$\begin{gathered} {\mathrm{Fe}}_3{\mathrm{O}}_4+\mathrm{HCl}\to {\mathrm{FeCl}}_2+{\mathrm{FeCl}}_3+{\mathrm{H}}_2\mathrm{O} \\ (\mathrm{Ferroso}-\mathrm{ferric}\mathrm{oxide})(\mathrm{Hydrochloric}\mathrm{acid})(\mathrm{Ferrous}\mathrm{chloride})(\mathrm{Ferric}\mathrm{chloride})\left(\mathrm{Water}\right) \end{gathered}$$

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.
Iron (Fe)	3	2
Oxygen (O)	4	1
Hydrogen (H)	1	2
Chlorine (Cl)	1	5

Balance the equation by multiplying Ferric chloride with 2 on RHS as the number of atoms of Iron are less on RHS.

$$\begin{gathered} {\mathrm{Fe}}_3{\mathrm{O}}_4+\mathrm{HCl}\to {\mathrm{FeCl}}_2+2{\mathrm{FeCl}}_3+{\mathrm{H}}_2\mathrm{O} \\ (\mathrm{Ferroso}-\mathrm{ferric}\mathrm{oxide})(\mathrm{Hydrochloric}\mathrm{acid})(\mathrm{Ferrous}\mathrm{chloride})(\mathrm{Ferric}\mathrm{chloride})\left(\mathrm{Water}\right) \end{gathered}$$

Type of atoms	Number of atoms on L.H.S.	Number of atoms on R.H.S.
Iron (Fe)	3	3
Oxygen (O)	4	1
Hydrogen (H)	1	2
Chlorine (Cl)	1	8

As the number of atoms are still unequal on both side, multiply hydrochloric acid with 2 on LHS as the number of atoms of chlorine are less on LHS.

$$\begin{gathered} {\mathrm{Fe}}_3{\mathrm{O}}_4+8\mathrm{HCl}\to {\mathrm{FeCl}}_2+2{\mathrm{FeCl}}_3+{\mathrm{H}}_2\mathrm{O} \\ (\mathrm{Ferroso}-\mathrm{ferric}\mathrm{oxide})(\mathrm{Hydrochloric}\mathrm{acid})(\mathrm{Ferrous}\mathrm{chloride})(\mathrm{Ferric}\mathrm{chloride})\left(\mathrm{Water}\right) \end{gathered}$$

Type of atoms	Number of atoms on L.H.S.	Number of atoms on R.H.S.
Iron (Fe)	3	3
Oxygen (O)	4	1
Hydrogen (H)	8	2
Chlorine (Cl)	8	8

As the number of atoms are still unequal on both side, multiply water with 4 on RHS as the number of atoms of oxygen are less on RHS.

$$\begin{gathered} {\mathrm{Fe}}_3{\mathrm{O}}_4+8\mathrm{HCl}\to {\mathrm{FeCl}}_2+2{\mathrm{FeCl}}_3+4{\mathrm{H}}_2\mathrm{O} \\ (\mathrm{Ferroso}-\mathrm{ferric}\mathrm{oxide})(\mathrm{Hydrochloric}\mathrm{acid})(\mathrm{Ferrous}\mathrm{chloride})(\mathrm{Ferric}\mathrm{chloride})\left(\mathrm{Water}\right) \end{gathered}$$

Type of atoms	Number of atoms on L.H.S.	Number of atoms on R.H.S.
Iron (Fe)	3	3
Oxygen (O)	4	4
Hydrogen (H)	8	8
Chlorine (Cl)	8	8

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

$$\begin{gathered} {\mathrm{Fe}}_3{\mathrm{O}}_4\left(\mathrm{s}\right)+8\mathrm{HCl}\left(\mathrm{l}\right)\to {\mathrm{FeCl}}_2\left(\mathrm{aq}\right)+2{\mathrm{FeCl}}_3\left(\mathrm{aq}\right)+4{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right) \\ (\mathrm{Ferroso}-\mathrm{ferric}\mathrm{oxide})(\mathrm{Hydrochloric}\mathrm{acid})(\mathrm{Ferrous}\mathrm{chloride})(\mathrm{Ferric}\mathrm{chloride})\left(\mathrm{Water}\right) \end{gathered}$$