# Calculating Oxygen Gas Release in Potassium Chlorate Decomposition

## How can we determine the volume of oxygen gas released at STP when potassium chlorate is decomposed?

Given the balanced equation for the reaction and the molar mass of KClO3, what is the volume of oxygen gas released when 10.0 g of potassium chlorate is decomposed?

### Options:

0.914 L

1.83 L

2.74 L

3.66 L

## Answer:

Third option 2.74 L

The balanced reaction for the decomposition of potassium chlorate is 2KClO3 → 2KCl + 3O2.

We are given the amount of potassium chlorate as 10.0 g. To calculate the volume of oxygen gas released, we first convert the given mass of KClO3 to moles, then use the molar ratio from the balanced equation to find the moles of O2 produced, and finally apply the ideal gas law formula to determine the volume at STP.

Calculation:

10.0 g KClO3 x (1 mol KClO3 / 122.55 g KClO3) x (3 mol O2 / 2 mol KClO3) = 0.1224 mol O2

Volume (V) = nRT/P = 0.1224 mol O2 x 273 K x 0.08206 atm L/mol K / 1 atm = 2.74 L

Therefore, the volume of oxygen gas released at STP when 10.0 g of potassium chlorate is decomposed is 2.74 L.