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To determine how many moles of [tex]\( O_2 \)[/tex] are produced when 0.50 mol of [tex]\( O_3(g) \)[/tex] reacts, let's follow these steps:
1. Start with the balanced chemical equation:
[tex]\[ 2 O_3(g) \rightarrow 3 O_2(g) \][/tex]
This equation tells us that 2 moles of ozone ([tex]\( O_3 \)[/tex]) produce 3 moles of oxygen gas ([tex]\( O_2 \)[/tex]).
2. Establish the mole ratio from the balanced equation:
The mole ratio of [tex]\( O_3 \)[/tex] to [tex]\( O_2 \)[/tex] from the balanced equation is 2:3. This means that for every 2 moles of [tex]\( O_3 \)[/tex] consumed, 3 moles of [tex]\( O_2 \)[/tex] are produced.
3. Calculate the moles of [tex]\( O_2 \)[/tex] produced:
We are given that 0.50 mol of [tex]\( O_3 \)[/tex] reacts. Using the mole ratio, we can determine the moles of [tex]\( O_2 \)[/tex] produced:
[tex]\[ \text{Moles of } O_2 = \left(\frac{3 \text{ moles of } O_2}{2 \text{ moles of } O_3}\right) \times 0.50 \text{ moles of } O_3 \][/tex]
Simplifying the expression:
[tex]\[ \text{Moles of } O_2 = \frac{3}{2} \times 0.50 \][/tex]
[tex]\[ \text{Moles of } O_2 = 1.5 \times 0.50 \][/tex]
[tex]\[ \text{Moles of } O_2 = 0.75 \][/tex]
Therefore, when 0.50 mol of [tex]\( O_3(g) \)[/tex] reacts, 0.75 mol of [tex]\( O_2(g) \)[/tex] are produced.
The correct answer is C. 0.75.
1. Start with the balanced chemical equation:
[tex]\[ 2 O_3(g) \rightarrow 3 O_2(g) \][/tex]
This equation tells us that 2 moles of ozone ([tex]\( O_3 \)[/tex]) produce 3 moles of oxygen gas ([tex]\( O_2 \)[/tex]).
2. Establish the mole ratio from the balanced equation:
The mole ratio of [tex]\( O_3 \)[/tex] to [tex]\( O_2 \)[/tex] from the balanced equation is 2:3. This means that for every 2 moles of [tex]\( O_3 \)[/tex] consumed, 3 moles of [tex]\( O_2 \)[/tex] are produced.
3. Calculate the moles of [tex]\( O_2 \)[/tex] produced:
We are given that 0.50 mol of [tex]\( O_3 \)[/tex] reacts. Using the mole ratio, we can determine the moles of [tex]\( O_2 \)[/tex] produced:
[tex]\[ \text{Moles of } O_2 = \left(\frac{3 \text{ moles of } O_2}{2 \text{ moles of } O_3}\right) \times 0.50 \text{ moles of } O_3 \][/tex]
Simplifying the expression:
[tex]\[ \text{Moles of } O_2 = \frac{3}{2} \times 0.50 \][/tex]
[tex]\[ \text{Moles of } O_2 = 1.5 \times 0.50 \][/tex]
[tex]\[ \text{Moles of } O_2 = 0.75 \][/tex]
Therefore, when 0.50 mol of [tex]\( O_3(g) \)[/tex] reacts, 0.75 mol of [tex]\( O_2(g) \)[/tex] are produced.
The correct answer is C. 0.75.
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