Final answer:
By using stoichiometry and the balanced equation 2KClO₃ → 2KCl + 3O₂, we calculate that 36.768 grams of O₂ can be theoretically produced from 93.9 grams of KClO₃.
Step-by-step explanation:
To calculate how many grams of O₂ can be produced by heating 93.9 grams of KClO₃, we'll start with a stoichiometric calculation based on the balanced chemical equation provided: 2KClO₃ (s) → 2KCl (s) + 3O₂ (g). First, the molar mass of KClO₃ is needed to convert the mass of KClO₃ to moles.
1 mol KClO₃ = 122.55 g/mol
1 mol O₂ = 32.00 g/mol
Using these molar masses, the moles of KClO₃ involved in the reaction can be calculated:
93.9 g KClO₃ × (1 mol KClO₃ / 122.55 g KClO₃) = 0.766 moles of KClO₃
From the balanced equation, we see that 2 moles of KClO₃ produce 3 moles of O₂. So, we can calculate the moles of O₂ produced from the moles of KClO₃:
0.766 moles KClO₃ × (3 moles O₂ / 2 moles KClO₃) = 1.149 moles O₂
Finally, this result can be converted to grams of O₂:
1.149 moles O₂ × (32.00 g O₂ / 1 mol O₂) = 36.768 g O₂
Thus, you can theoretically produce 36.768 grams of O₂ from heating 93.9 grams of KClO₃.