# 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!