Question

Iron oxide ores, commonly a mixture of$\mathbf{FeO}$and ${\mathbf{Fe}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}$are given the general formula ${\mathbf{Fe}}_{\mathbf{3}}{\mathbf{O}}_{\mathbf{4}}$.
They yield elemental iron when heated to a very high temperature with either carbon monoxideor elemental hydrogen. Balance the following equations for these processes:

${\mathbf{Fe}}_{\mathbf{3}}{\mathbf{O}}_{\mathbf{4}}\left(\mathbf{s}\right)\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\left(\mathbf{g}\right)\to \mathbf{Fe}\left(\mathbf{s}\right)\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(\mathbf{g}\right)$

${\mathbf{Fe}}_{\mathbf{3}}{\mathbf{O}}_{\mathbf{4}}\left(\mathbf{s}\right)\mathbf{+}\mathbf{CO}\left(\mathbf{g}\right)\to \mathbf{Fe}\left(\mathbf{s}\right)\mathbf{+}{\mathbf{CO}}_{\mathbf{2}}\left(\mathbf{g}\right)$

Short answer

The following equations yield elemental iron when heated to a very high temperature with either carbon monoxide or elemental hydrogen:

${\text{Fe}}_3{\text{O}}_4\left(s\right)+4H_2\left(g\right)\to 3Fe\left(s\right)+4H_{2}O\left(g\right)$and

${\text{Fe}}_3{\text{O}}_4\left(s\right)+4CO\left(g\right)\to 3Fe\left(s\right)+4C{O}_2\left(g\right)$

Step-by-step solution

Definition

A balanced chemical equation is one in which the number of atoms present on the reactant side of the equation equals the number of atoms present on the product side of the equation.

Balancing reactant side chemical equation of first equation

The first equation is ${\text{Fe}}_3{\text{O}}_4\left(s\right)+H_2\left(g\right)\to Fe\left(s\right)+H_{2}O\left(g\right)$

On the reactant side, there are three $\text{Fe}$atoms. So, first, balance the $\text{Fe}$atoms by assigning a coefficient 3 to $\text{Fe}$of products, as illustrated below:

${\text{Fe}}_3{\text{O}}_4\left(s\right)+H_2\left(g\right)\to 3Fe\left(s\right)+H_{2}O\left(g\right)$

Balanced chemical equation of first equation

The oxygen atoms can then be balanced by adding a coefficient 4 to the $H_{2}O$of the product, yielding the following equation:

${\text{Fe}}_3{\text{O}}_4\left(s\right)+4H_2\left(g\right)\to 3Fe\left(s\right)+4H_{2}O\left(g\right)$

So, products have an average of 8 hydrogen atoms. As a result, the reactant $H_2$can be given a coefficient of 4. Thus, the following is the final balanced chemical equation:

${\text{Fe}}_3{\text{O}}_4\left(s\right)+4H_2\left(g\right)\to 3Fe\left(s\right)+4H_{2}O\left(g\right)$

Balancing reactant side chemical equation of second equation

The second equation is ${\text{Fe}}_3{\text{O}}_4\left(s\right)+CO\left(g\right)\to Fe\left(s\right)+C{O}_2\left(g\right)$

On the reactant side, there are three $\text{Fe}$atoms. So, first, balance $\text{Fe}$the atoms by assigning a coefficient 3 to $\text{Fe}$of products, as illustrated below:

${\text{Fe}}_3{\text{O}}_4\left(s\right)+CO\left(g\right)\to 3Fe\left(s\right)+C{O}_2\left(g\right)$

Balanced chemical equation of second equation

After that, the oxygen atoms can be balanced by applying a coefficient of 4 to both the$\text{CO}$of the reactant and the${\text{CO}}_2$of the product, yielding the equation:

${\text{Fe}}_3{\text{O}}_4\left(s\right)+4CO\left(g\right)\to 3Fe\left(s\right)+4C{O}_2\left(g\right)$

The number of$\text{C}$atoms is also balanced as a result of this. So, the balanced chemical equation is as follows:

${\text{Fe}}_3{\text{O}}_4\left(s\right)+4CO\left(g\right)\to 3Fe\left(s\right)+4C{O}_2\left(g\right)$