Iron oxide ores, commonly a mixture of 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: $$ \begin{array}{c}{\mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{Fe}(s)+\mathrm{H}_{2} \mathrm{O}(g)} \\\ {\mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{CO}(g) \longrightarrow \mathrm{Fe}(s)+\mathrm{CO}_{2}(g)}\end{array} $$

Begin by listing the number of atoms for each element on both sides of the equation: Reactants: Fe (3), O (4), H (2) Products: Fe (1), O (1), H (2) To balance the iron atoms, we need 3 moles of Fe on the product side. So we'll add a coefficient of 3 in front of Fe(s) in the product side: Fe3O4(s) + H2(g) -> 3Fe(s) + H2O(g) Now the iron atoms are balanced, and we have: Reactants: Fe (3), O (4), H (2) Products: Fe (3), O (1), H (2) Next, we'll balance the oxygen atoms by placing a coefficient of 4 in front of H2O(g) in the product side: Fe3O4(s) + H2(g) -> 3Fe(s) + 4H2O(g) Now we have: Reactants: Fe (3), O (4), H (2) Products: Fe (3), O (4), H (8) Finally, we'll balance the hydrogen atoms by placing a coefficient of 4 in front of H2(g) in the reactant side: Fe3O4(s) + 4H2(g) -> 3Fe(s) + 4H2O(g) Now the reaction is balanced, with equal numbers of each atom on both sides.

Begin by listing the number of atoms for each element on both sides of the equation: Reactants: Fe (3), O (4), C (1), O (1) Products: Fe (1), C (1), O (2) To balance the iron atoms, we need 3 moles of Fe on the product side. So we'll add a coefficient of 3 in front of Fe(s) in the product side: Fe3O4(s) + CO(g) -> 3Fe(s) + CO2(g) Now the iron atoms are balanced, and we have: Reactants: Fe (3), O (4), C (1), O (1) Products: Fe (3), C (1), O (6) Next, we'll balance the carbon atoms by placing a coefficient of 3 in front of CO(g) in the reactant side: Fe3O4(s) + 3CO(g) -> 3Fe(s) + CO2(g) Now we have: Reactants: Fe (3), O (4), C (3), O (3) Products: Fe (3), C (1), O (6) Finally, we'll balance the oxygen atoms by placing a coefficient of 3 in front of CO2(g) in the product side: Fe3O4(s) + 3CO(g) -> 3Fe(s) + 3CO2(g) Now the reaction is balanced, with equal numbers of each atom on both sides.