Of course! Let's break this problem down, step by step, to determine how many moles of oxygen ([tex]\(O_2\)[/tex]) will be produced when 0.07 mol of potassium chlorate ([tex]\(KClO_3\)[/tex]) decomposes.
1. Identify the stoichiometry of the reaction:
The balanced chemical equation is:
[tex]\[
2 KClO_3(s) \rightarrow 3 O_2(g) + 2 KCl(s)
\][/tex]
This tells us that 2 moles of [tex]\(KClO_3\)[/tex] produce 3 moles of [tex]\(O_2\)[/tex].
2. Set up the mole ratio:
From the balanced equation, the mole ratio between [tex]\(KClO_3\)[/tex] and [tex]\(O_2\)[/tex] is:
[tex]\[
\text{Moles of } O_2 = \left(\frac{3 \text{ moles } O_2}{2 \text{ moles } KClO_3}\right)
\][/tex]
3. Calculate the moles of [tex]\(O_2\)[/tex] produced:
Given that 0.07 mol of [tex]\(KClO_3\)[/tex] decomposed, we can use the mole ratio to find the moles of [tex]\(O_2\)[/tex]:
[tex]\[
\text{Moles of } O_2 = 0.07 \text{ moles } KClO_3 \times \left(\frac{3 \text{ moles } O_2}{2 \text{ moles } KClO_3}\right)
\][/tex]
4. Simplify the expression:
[tex]\[
\text{Moles of } O_2 = 0.07 \times 1.5
\][/tex]
5. Perform the multiplication:
[tex]\[
\text{Moles of } O_2 = 0.105
\][/tex]
So, 0.105 moles of oxygen ([tex]\(O_2\)[/tex]) will be produced from the decomposition of 0.07 mol of potassium chlorate ([tex]\(KClO_3\)[/tex]).
