The reaction you provided involves iron(III) oxide (Fe2O3) reacting with sulfuric acid (H2SO4). The balanced reaction is:
\[ \text{Fe}_2\text{O}_3 + 3 \text{H}_2\text{SO}_4 \rightarrow \text{Fe}_2(\text{SO}_4)_3 + 3 \text{H}_2\text{O} \]
In this balanced equation:
- Each formula unit of Fe2O3 has 3 oxygen atoms.
- Each formula unit of H2SO4 has 4 oxygen atoms, and since there are 3 H2SO4, that contributes \( 3 \times 4 = 12 \) oxygen atoms.
In the products, the sulfate (Fe2(SO4)3) contains \( 3 \times 4 = 12 \) oxygen atoms, and the water adds \( 3 \) (1 from each H2O).
So, in the products we have:
- 12 oxygen atoms from \( \text{Fe}_2(\text{SO}_4)_3 \)
- 3 oxygen atoms from \( 3 \text{H}_2\text{O} \)
Total oxygen atoms in products = \( 12 + 3 = 15 \).
Thus, the answer is \( 15 \).