Chemistry Fun: Calculating Oxygen Gas Volume Released
How can we calculate the volume of oxygen gas released at STP when 10.0 g of potassium chlorate is decomposed?
Given the molar mass of KClO₃ is 122.55 g/mol and the balanced equation for the reaction.
Answer:
Your answer is 2.74.
To calculate the volume of oxygen gas released at STP when 10.0 g of potassium chlorate is decomposed, we first need to determine the number of moles of oxygen produced. According to the balanced equation given:
2 KClO₃ → 2 KCl + 3 O₂
We start by finding the mass of oxygen produced by using the given mass of potassium chlorate and the molar mass of oxygen (O₂).
Given data:
Molar mass:
KClO₃ = 122.55 g/mol
O₂ = 32 g/mol
Mass of KClO₃ = 10.0 g
Now, we can calculate the mass of oxygen produced:
Mass of O₂ = (10.0 g × 3 × 32 g) / (2 × 122.55)
Mass of O₂ = 960 / 245.1
Mass of O₂ = 3.9167 g
Next, we calculate the number of moles of oxygen produced:
Number of moles O₂ = 3.9167 g / 32 g/mol = 0.1223 moles
Finally, we can determine the volume of oxygen gas produced at STP:
1 mole of gas at STP occupies 22.4 L
Volume of O₂ = 0.1223 moles × 22.4 L/mol = 2.739 L
Therefore, the volume of oxygen gas released at STP when 10.0 g of potassium chlorate is decomposed is 2.739 liters. Chemistry can be fun and exciting when you see the calculations come to life!