Answer:
approximately 4.8 grams of oxygen gas is formed.
Step-by-step explanation:
To determine the amount of oxygen (O2) formed when 12.26 grams of potassium chlorate (KClO3) is heated, we need to consider the chemical reaction that occurs during this process. When KClO3 is heated, it decomposes into potassium chloride (KCl) and oxygen gas (O2).
The balanced chemical equation for this reaction is:
2 KClO3(s) → 2 KCl(s) + 3 O2(g)
From the equation, we can see that 2 moles of KClO3 produce 3 moles of O2.
Calculate the molar mass of KClO3:
K: 39.10 g/mol
Cl: 35.45 g/mol
O: 16.00 g/mol
Molar mass of KClO3 = (39.10 + 35.45 + 3 * 16.00) g/mol = 122.55 g/mol
Calculate the number of moles of KClO3 in 12.26 grams:
Moles of KClO3 = (12.26 g) / (122.55 g/mol) = 0.1 moles
Now, we can determine the number of moles of O2 produced using the mole ratio from the balanced equation:
Moles of O2 = (0.1 moles of KClO3) * (3 moles of O2 / 2 moles of KClO3) = 0.15 moles
Finally, convert moles of O2 to grams:
Mass of O2 = (0.15 moles) * (32.00 g/mol for O2) = 4.8 grams
So, when 12.26 grams of KClO3 is heated, approximately 4.8 grams of oxygen gas is formed.