Decomposition of potassium chlorate (KClO3) produces potassium chloride (KCl) and pure oxygen (O2). The balanced equation for the reaction is as follows: 2KClO3(s) —> 2KCl(s) + 3O2(g). What volume of oxygen gas is released at STP if 10.0 g of potassium chlorate is decomposed? (The molar mass of KClO3 is 122.55 g/mol.)
The correct answer and explanation is :
To determine the volume of oxygen gas (O₂) released at standard temperature and pressure (STP) when 10.0 g of potassium chlorate (KClO₃) decomposes, follow these steps:
Step 1: Calculate the Moles of KClO₃
Given:
- Mass of KClO₃ = 10.0 g
- Molar mass of KClO₃ = 122.55 g/mol
Moles of KClO₃:
[
\frac{10.0 \text{ g}}{122.55 \text{ g/mol}} = 0.0816 \text{ moles}
]
Step 2: Use the Balanced Equation to Find Moles of O₂
From the reaction:
[
2KClO₃ \rightarrow 2KCl + 3O₂
]
The molar ratio of KClO₃ to O₂ is 2:3. Using this ratio:
[
0.0816 \text{ moles KClO₃} \times \frac{3 \text{ moles O₂}}{2 \text{ moles KClO₃}}
]
[
= 0.1224 \text{ moles O₂}
]
Step 3: Convert Moles of O₂ to Volume at STP
At STP, 1 mole of any gas occupies 22.4 L.
[
0.1224 \text{ moles O₂} \times 22.4 \text{ L/mole}
]
[
= 2.74 \text{ L O₂}
]
Final Answer:
The volume of oxygen gas released is 2.74 L at STP.
Explanation:
This problem involves stoichiometric calculations using the balanced chemical equation for potassium chlorate decomposition. The key steps include:
- Converting mass of reactant (KClO₃) to moles using molar mass.
- Using the balanced equation to find the stoichiometric relationship between KClO₃ and O₂.
- Converting moles of O₂ to volume using the molar volume of a gas at STP.
This method ensures accurate results by following a structured approach to chemical calculations.
