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Taking into account the reaction stoichiometry, 9717.75 grams of oxygen gas are produced when 2.43×10⁴ grams of KClO₃ are completely reacted.
The balanced reaction is:
2 KClO₃ → 2 KCl + 3 O₂
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- KClO₃: 2 moles
- KCl: 2 moles
- O₂: 3 moles
The molar mass of each compound is:
- KClO₃: 122.55 g/mole
- KCl: 74.55 g/mole
- O₂: 32 g/mole
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- KClO₃: 2 moles× 122.55 g/mole= 245.1 grams
- KCl: 2 moles× 74.55 g/mole= 149.1 grams
- O₂: 3 moles× 32 g/mole= 96 grams
Then you can apply the following rule of three: if by reaction stoichiometry 245.1 grams of KClO₃ form 96 grams of O₂, 2.43×10⁴ grams of KClO₃ form how much mass of O₂?
[tex]mass of O_{2} =\frac{2.43x10^{4} grams of KClO_{3} x 96 grams of O_{2}}{245.1grams of KClO_{3} }[/tex]
mass of O₂= 9517.75 grams
In summary, 9717.75 grams of oxygen gas are produced when 2.43×10⁴ grams of KClO₃ are completely reacted.
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