To figure out how many moles of [tex]\( O_2 \)[/tex] are produced when 0.50 mol of [tex]\( O_3 \)[/tex] reacts, we'll use the stoichiometric relationship given in the balanced chemical equation:
[tex]\[ 2 O_3(g) \rightarrow 3 O_2(g) \][/tex]
This tells us that 2 moles of [tex]\( O_3 \)[/tex] produce 3 moles of [tex]\( O_2 \)[/tex]. We can set up a ratio to find out how many moles of [tex]\( O_2 \)[/tex] are produced from the 0.50 mol of [tex]\( O_3 \)[/tex] we have:
1. First, determine the mole ratio between [tex]\( O_2 \)[/tex] and [tex]\( O_3 \)[/tex]:
[tex]\[
\text{Molar ratio } = \frac{3 \text{ moles } O_2}{2 \text{ moles } O_3}
\][/tex]
2. Use this ratio to find the moles of [tex]\( O_2 \)[/tex] produced from 0.50 mol of [tex]\( O_3 \)[/tex]:
[tex]\[
\text{Moles of } O_2 = 0.50 \text{ mol } O_3 \times \frac{3 \text{ mol } O_2}{2 \text{ mol } O_3}
\][/tex]
3. Multiply the values:
[tex]\[
\text{Moles of } O_2 = 0.50 \times 1.5 = 0.75
\][/tex]
Therefore, 0.75 moles of [tex]\( O_2 \)[/tex] are produced when 0.50 mol of [tex]\( O_3 \)[/tex] reacts.
Hence, the correct answer is:
A. 0.75
