Chemistry · Chemical equations and stoichiometry
On heating at a certain temperature, it is observed that one mole of yields one
On heating \( K C l O_{3} \) at a certain temperature, it is observed that one mole of \( K C l O_{3} \) yields one mole of \( O_{2} \) What is the mole fraction of \( K C l O_{4} \) in the final solid mixture containing only \( K C l \) and \( K C l O_{4}, \) the latter being formed by the parallel reaction?
- A. 0.50
- B. 0.25
- C. 0.75
- D. 0.67
Step-by-step solution
Let x be the fraction of KClO3 decomposing via 2KClO3 → 2KCl + 3O2. Since 1 mole KClO3 yields 1 mole O2, and reaction (1) gives 1.5 moles O2 per mole, we have 1.5x = 1 → x = 2/3, so fraction via (2) is 1/3. Reaction (2): 4KClO3 → 3KClO4 + KCl gives 0.75 mole KClO4 per mole. Total KClO4 = (1/3)*0.75 = 0.25. Total solids: KCl from (1): 2/3, from (2): (1/3)*0.25 = 1/12, sum = 3/4; plus KClO4: 0.25; total = 1. Mole fraction KClO4 = 0.25.
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