To balance the reaction \( 4 \text{Al} + _ \text{O}_2 \rightarrow 2 \text{Al}_2\text{O}_3 \), we first examine the products.
Each formula unit of \( \text{Al}_2\text{O}_3 \) contains 2 aluminum (Al) atoms and 3 oxygen (O) atoms. Since there are 2 formula units of \( \text{Al}_2\text{O}_3 \) in the reaction, we have:
- Total aluminum in products: \( 2 \times 2 = 4 \) Al (which is balanced with the 4 Al on the reactant side).
- Total oxygen in products: \( 2 \times 3 = 6 \) O.
Now we look at the reactants. We need a total of 6 oxygen atoms, and since \( \text{O}_2 \) provides 2 oxygen atoms per molecule, we need:
\[ \frac{6 \text{ O}}{2 \text{ O}_2} = 3 \text{ O}_2 \]
So, the coefficient that should be placed in front of \( \text{O}_2 \) is 3.
Thus, the balanced reaction is:
\[ 4 \text{Al} + 3 \text{O}_2 \rightarrow 2 \text{Al}_2\text{O}_3 \]
The answer is 3.