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How many grams of KClO₃ were in the original mixture?

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Final answer:

To find the grams of KClO₃ in the original mixture, use the mass of collected O₂ to perform stoichiometric calculations based on the molar masses of O₂ and KClO₃, and the balanced chemical equation.

Step-by-step explanation:

The student wants to know how many grams of KClO₃ were in the original mixture. To find this, we need to calculate the theoretical yield of oxygen gas (O₂) and then use stoichiometry to compute the amount of KClO₃ that would produce this yield of O₂.

Given: Mass of O₂ collected = 14.9 g. According to the problem, 1 mol of O₂ = 32.00 g/mol. The balanced equation requires 8 mol of KClO₃ to react with 1 mol of C12H22O11, making KClO₃ the limiting reactant. Therefore, we use the molar mass of KClO₃ (122.55 g/mol) to find the mass of KClO₃ that produced the given mass of O₂.

The calculation would be 14.9 g O₂ x (1 mol O₂/32.00 g O₂) x (8 mol KClO₃/3 mol O₂) x (122.55 g KClO₃/1 mol KClO₃). By solving, we find that the mass of KClO₃ in the original mixture is the answer to this calculation.

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