Potassium chlorate decomposes to potassium chloride and oxygen. If 20.8 g of potassium chlorate decomposes, how many liters of oxygen will form at STP?

5.71 L of oxygen will be formed at STP

Explanation:

Step 1: Data given

Mass of potassium chlorate = 20.8 grams

Molar mass potassium chlorate = 122.55 g/mol

Step 2: The balanced equation

2KClO3 → 2KCl + 3O2

Step 3: Calculate moles KClO3

Moles KClO3 = mass KClO3 / molar mass KClO3

Moles KClO3 = 20.8 grams / 122.55 g/mol

Moles KClO3 = 0.170 moles

Step 4: Calculate moles O2

For 2 moles KClO3 we'll have 2 moles KCl and 3 moles O2

For 0.170 moles KClO3 we'll have 3/2 *0.170 = 0.255 moles O2

Step 5: Calculate volume O2

1 mol = 22.4 L

0.255 moles = 5.71 L

5.71 L of oxygen will be formed at STP

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