Balance this chemical equation by inspection: \(\mathrm{Fe}_{2} \mathrm{O}_{3}+\mathrm{C} \rightarrow \mathrm{Fe}+\mathrm{CO}_{2}\)

In this exercise, the given chemical equation is \[\mathrm{Fe}_{2} \mathrm{O}_{3}+\mathrm{C} \rightarrow \mathrm{Fe}+\mathrm{CO}_{2}.\] The reactants are \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) (iron(III) oxide) and \(\mathrm{C}\) (carbon), and the products are \(\mathrm{Fe}\) (iron) and \(\mathrm{CO}_{2}\) (carbon dioxide).

For this step, let's make a table to keep track of the number of atoms for each element. Reactant side: \[\mathrm{Fe}: 2 \quad \mathrm{O}: 3 \quad \mathrm{C}: 1\] Product side: \[\mathrm{Fe}: 1 \quad \mathrm{O}: 2 \quad \mathrm{C}: 1\]

Now, let's balance the atoms one at a time. Begin with the element that appears the fewest times in the equation, which is iron (Fe). Reactant side: \[\mathrm{Fe}: 2 \quad \mathrm{O}: 3 \quad \mathrm{C}: 1\] Product side: \[\mathrm{Fe}: 2 \quad \mathrm{O}: 2 \quad \mathrm{C}: 1\] We can achieve balancing iron by multiplying iron (Fe) on the product side by 2. Now, the equation looks like this: \[\mathrm{Fe}_{2} \mathrm{O}_{3}+\mathrm{C} \rightarrow 2\mathrm{Fe}+\mathrm{CO}_{2}.\] Next, let's move on to balance the atoms of oxygen (O). To balance the oxygen atoms, we can multiply carbon dioxide (CO2) by 3: \[\mathrm{Fe}_{2} \mathrm{O}_{3}+\mathrm{C} \rightarrow 2\mathrm{Fe}+3\mathrm{CO}_{2}.\] Reactant side: \[\mathrm{Fe}: 2 \quad \mathrm{O}: 3 \quad \mathrm{C}: 1\] Product side: \[\mathrm{Fe}: 2 \quad \mathrm{O}: 6 \quad \mathrm{C}: 3\] Finally, let's balance the atoms of carbon (C). We can see that there are 3 carbon atoms on the product side and only 1 on the reactant side. So, we can multiply the carbon atom (C) on the reactant side by 3: \[3(\mathrm{Fe}_{2} \mathrm{O}_{3}+\mathrm{C}) \rightarrow 2\mathrm{Fe}+3\mathrm{CO}_{2}.\] Reactant side: \[\mathrm{Fe}: 2 \quad \mathrm{O}: 3 \quad \mathrm{C}: 3\] Product side: \[\mathrm{Fe}: 2 \quad \mathrm{O}: 6 \quad \mathrm{C}: 3\] The equation is now balanced.

The balanced chemical equation is: \[3\mathrm{C} + 2\mathrm{Fe}_{2} \mathrm{O}_{3} \rightarrow 4\mathrm{Fe} + 3\mathrm{CO}_{2}\]