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Decomposition of potassium chlorate ([tex]$KClO_3$[/tex]) produces potassium chloride ([tex]$KCl$[/tex]) and pure oxygen ([tex]$O_2$[/tex]). The balanced equation for the reaction is as follows:

[tex]\[2 KClO_3(s) \longrightarrow 2 KCl(s) + 3 O_2(g)\][/tex]

What volume of oxygen gas is released at STP if 10.0 g of potassium chlorate is decomposed? (The molar mass of [tex]$KClO_3$[/tex] is [tex]$122.55 \, g/mol$[/tex])

A. [tex]$0.914 \, L$[/tex]
B. [tex]$1.83 \, L$[/tex]
C. [tex]$2.74 \, L$[/tex]
D. [tex]$3.66 \, L$[/tex]

Sagot :

GinnyAnswer
To determine the volume of oxygen gas released at STP (Standard Temperature and Pressure: [tex]\(0^\circ\text{C} \)[/tex] and [tex]\(1 \text{atm}\)[/tex]) when [tex]\(10.0\)[/tex] grams of potassium chlorate ([tex]\( KClO_3 \)[/tex]) is decomposed, we'll follow a step-by-step process.

### 1. Calculation of Moles of [tex]\( KClO_3 \)[/tex]
We first need to determine the number of moles of [tex]\( KClO_3 \)[/tex] decomposed.

Given:
- Mass of [tex]\( KClO_3 \)[/tex]: [tex]\( 10.0 \text{ g} \)[/tex]
- Molar mass of [tex]\( KClO_3 \)[/tex]: [tex]\( 122.55 \text{ g/mol} \)[/tex]

The number of moles is given by:
[tex]\[ \text{moles of } KClO_3 = \frac{\text{mass of } KClO_3}{\text{molar mass of } KClO_3} \][/tex]

[tex]\[ \text{moles of } KClO_3 = \frac{10.0 \text{ g}}{122.55 \text{ g/mol}} \approx 0.0816 \text{ mol} \][/tex]

### 2. Use the Balanced Equation to Find Moles of [tex]\( O_2 \)[/tex]
From the balanced chemical equation:
[tex]\[ 2 KClO_3 (s) \rightarrow 2 KCl (s) + 3 O_2 (g) \][/tex]

This equation tells us that 2 moles of [tex]\( KClO_3 \)[/tex] produce 3 moles of [tex]\( O_2 \)[/tex].

Therefore, the moles of [tex]\( O_2 \)[/tex] produced can be calculated as follows:
[tex]\[ \text{moles of } O_2 = \left( \frac{3}{2} \right) \times \text{moles of } KClO_3 \][/tex]

Substituting the moles of [tex]\( KClO_3 \)[/tex]:
[tex]\[ \text{moles of } O_2 = \left( \frac{3}{2} \right) \times 0.0816 \text{ mol} \approx 0.1224 \text{ mol} \][/tex]

### 3. Convert Moles of [tex]\( O_2 \)[/tex] to Volume at STP
At STP, 1 mole of any gas occupies [tex]\(22.414 \text{ L}\)[/tex].

Therefore, the volume of [tex]\( O_2 \)[/tex] can be calculated by:
[tex]\[ \text{Volume of } O_2 = \text{moles of } O_2 \times 22.414 \text{ L/mol} \][/tex]

Substituting the moles of [tex]\( O_2 \)[/tex]:
[tex]\[ \text{Volume of } O_2 = 0.1224 \text{ mol} \times 22.414 \text{ L/mol} \approx 2.74 \text{ L} \][/tex]

### Conclusion
The volume of oxygen gas released at STP when [tex]\(10.0\)[/tex] grams of potassium chlorate is decomposed is approximately [tex]\(2.74 \text{ L}\)[/tex].

Therefore, the correct answer is:
[tex]\[ \boxed{2.74 \text{ L}} \][/tex]
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