Answer :
Final answer:
Approximately 0.632 moles of oxygen gas form when 51.7 g of potassium chlorate decomposes completely.
Explanation:
To determine how many moles of oxygen form when 51.7 g of potassium chlorate (KClO3) decomposes completely, we need to use the molar mass of KClO3 and the balanced chemical equation for the decomposition reaction.
The molar mass of KClO3 is 122.55 g/mol. From the balanced chemical equation, we know that 2 moles of KClO3 decompose to form 3 moles of oxygen gas, so the molar ratio is 2:3.
To calculate the number of moles of oxygen gas formed, we can use the following equation:
moles of KClO3 = mass of KClO3 / molar mass of KClO3
moles of oxygen gas = moles of KClO3 * (3 moles of oxygen gas / 2 moles of KClO3)
Substituting the given values into the equation, we have:
moles of KClO3 = 51.7 g / 122.55 g/mol ≈ 0.4213 mol
moles of oxygen gas = 0.4213 mol * (3 moles of oxygen gas / 2 moles of KClO3) ≈ 0.632 moles
Therefore, approximately 0.632 moles of oxygen gas form when 51.7 g of potassium chlorate decomposes completely.
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