Question
he equation represents the breakdown of potassium chlorate (KClO3)
2KClO3 --> 2KCl + 3O2
How many moles of oxygen can be produced from 0.16 moles of potassium chlorate?
Responses
0.48 mol
0.48 mol
0.11 mol
0.11 mol
0.24 mol
0.24 mol
0.34 mol
2KClO3 --> 2KCl + 3O2
How many moles of oxygen can be produced from 0.16 moles of potassium chlorate?
Responses
0.48 mol
0.48 mol
0.11 mol
0.11 mol
0.24 mol
0.24 mol
0.34 mol
Answers
GPT-4o mini
To determine how many moles of oxygen (\(O_2\)) can be produced from 0.16 moles of potassium chlorate (\(KClO_3\)), we can use the stoichiometry of the balanced chemical equation:
\[
2KClO_3 \rightarrow 2KCl + 3O_2
\]
From the equation, we see that 2 moles of \(KClO_3\) produce 3 moles of \(O_2\). We can set up a relationship to find out how many moles of \(O_2\) are produced from 0.16 moles of \(KClO_3\):
\[
\text{Moles of } O_2 = \frac{3 \text{ moles } O_2}{2 \text{ moles } KClO_3} \times \text{Moles of } KClO_3
\]
Plugging in 0.16 moles of \(KClO_3\):
\[
\text{Moles of } O_2 = \frac{3}{2} \times 0.16 = 0.24 \text{ moles } O_2
\]
Therefore, the answer is:
**0.24 mol**
\[
2KClO_3 \rightarrow 2KCl + 3O_2
\]
From the equation, we see that 2 moles of \(KClO_3\) produce 3 moles of \(O_2\). We can set up a relationship to find out how many moles of \(O_2\) are produced from 0.16 moles of \(KClO_3\):
\[
\text{Moles of } O_2 = \frac{3 \text{ moles } O_2}{2 \text{ moles } KClO_3} \times \text{Moles of } KClO_3
\]
Plugging in 0.16 moles of \(KClO_3\):
\[
\text{Moles of } O_2 = \frac{3}{2} \times 0.16 = 0.24 \text{ moles } O_2
\]
Therefore, the answer is:
**0.24 mol**