Iron oxide ores, commonly a mixture of \(\mathrm{FeO}\) and \(\mathrm{Fe}_{2} \mathrm{O}_{3}\), are given the general formula \(\mathrm{Fe}_{3} \mathrm{O}_{4}\). They yield elemental iron when heated to a very high temperature with either carbon monoxide or elemental hydrogen. Balance the following equations for these processes: $$ \begin{aligned} \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{H}_{2}(g) & \longrightarrow \mathrm{Fe}(s)+\mathrm{H}_{2} \mathrm{O}(g) \\ \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{CO}(g) & \longrightarrow \mathrm{Fe}(s)+\mathrm{CO}_{2}(g) \end{aligned} $$

Currently, the equation is: $$ \mathrm{Fe}_3\mathrm{O}_4(s) + \mathrm{H}_2(g) \longrightarrow \mathrm{Fe}(s) + \mathrm{H}_2\mathrm{O}(g) $$ We have 3 Fe atoms on the left-hand side and only 1 Fe atom on the right-hand side. To balance this, we add a coefficient of 3 in front of the elemental iron: $$ \mathrm{Fe}_3\mathrm{O}_4(s) + \mathrm{H}_2(g) \longrightarrow 3\mathrm{Fe}(s) + \mathrm{H}_2\mathrm{O}(g) $$

Now, we must balance the O and H atoms. We have 4 O atoms on the left-hand side and only 1 on the right-hand side. Let's add a coefficient of 4 in front of the water molecule: $$ \mathrm{Fe}_3\mathrm{O}_4(s) + \mathrm{H}_2(g) \longrightarrow 3\mathrm{Fe}(s) + 4\mathrm{H}_2\mathrm{O}(g) $$ We now have 4 O atoms on both sides. Finally, we need to balance the H atoms. Since there are 8 H atoms in the water molecules on the right-hand side, we add a coefficient of 4 in front of the hydrogen gas: $$ \mathrm{Fe}_3\mathrm{O}_4(s) + 4\mathrm{H}_2(g) \longrightarrow 3\mathrm{Fe}(s) + 4\mathrm{H}_2\mathrm{O}(g) $$ The balanced equation using hydrogen as the reducing agent is: $$ \mathrm{Fe}_3\mathrm{O}_4(s) + 4\mathrm{H}_2(g) \longrightarrow 3\mathrm{Fe}(s) + 4\mathrm{H}_2\mathrm{O}(g) $$ Balancing the equation with carbon monoxide as the reducing agent:

As before, we balance the iron atoms first: $$ \mathrm{Fe}_3\mathrm{O}_4(s) + \mathrm{CO}(g) \longrightarrow 3\mathrm{Fe}(s) + \mathrm{CO}_2(g) $$

Now, we must balance the O and C atoms. We have 4 O atoms on the left-hand side and 2 on the right-hand side in carbon dioxide. Let's add a coefficient of 2 in front of the carbon dioxide molecule: $$ \mathrm{Fe}_3\mathrm{O}_4(s) + \mathrm{CO}(g) \longrightarrow 3\mathrm{Fe}(s) + 2\mathrm{CO}_2(g) $$ We now have 4 O atoms on both sides. Finally, we need to balance the C atoms. Since there are 2 C atoms in the carbon dioxide molecules on the right-hand side, we add a coefficient of 2 in front of the carbon monoxide gas: $$ \mathrm{Fe}_3\mathrm{O}_4(s) + 2\mathrm{CO}(g) \longrightarrow 3\mathrm{Fe}(s) + 2\mathrm{CO}_2(g) $$ The balanced equation using carbon monoxide as the reducing agent is: $$ \mathrm{Fe}_3\mathrm{O}_4(s) + 2\mathrm{CO}(g) \longrightarrow 3\mathrm{Fe}(s) + 2\mathrm{CO}_2(g) $$