Balance the chemical equation below using the smallest possible whole number stoichiometric coefficients. \( \mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{~s}) \rightarrow \mathrm{Fe}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g}) \)

To balance the chemical equation \( \mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{~s}) \rightarrow \mathrm{Fe}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g}) \), we will follow these steps: 1. **Identify the number of atoms of each element on both sides of the equation.** - On the left side (reactants), we have: - Iron (Fe): 2 (from \( \mathrm{Fe}_{2} \mathrm{O}_{3} \)) - Oxygen (O): 3 (from \( \mathrm{Fe}_{2} \mathrm{O}_{3} \)) - On the right side (products), we have: - Iron (Fe): 1 (from \( \mathrm{Fe} \)) - Oxygen (O): 2 (from \( \mathrm{O}_{2} \)) 2. **Balance the iron (Fe) atoms first.** - Since there are 2 Fe atoms on the left, we need 2 Fe atoms on the right. We can do this by placing a coefficient of 2 in front of \( \mathrm{Fe} \): \[ \mathrm{Fe}_{2} \mathrm{O}_{3} \rightarrow 2 \mathrm{Fe} + \mathrm{O}_{2} \] 3. **Now, balance the oxygen (O) atoms.** - On the left, we have 3 O atoms, and on the right, we have 2 O atoms from \( \mathrm{O}_{2} \). To balance the oxygen, we can place a coefficient of \( \frac{3}{2} \) in front of \( \mathrm{O}_{2} \): \[ \mathrm{Fe}_{2} \mathrm{O}_{3} \rightarrow 2 \mathrm{Fe} + \frac{3}{2} \mathrm{O}_{2} \] 4. **Convert to whole number coefficients.** - To eliminate the fraction, we can multiply the entire equation by 2: \[ 2 \mathrm{Fe}_{2} \mathrm{O}_{3} \rightarrow 4 \mathrm{Fe} + 3 \mathrm{O}_{2} \] Now, the balanced equation with the smallest possible whole number stoichiometric coefficients is: \[ \boxed{2 \mathrm{Fe}_{2} \mathrm{O}_{3} \rightarrow 4 \mathrm{Fe} + 3 \mathrm{O}_{2}} \]