Answer:
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
