Oxygen can be prepared by the decomposition of potassium chlorate (KClO3). How many moles of oxygen (O2(g)) can be formed by taking 12 moles of potassium chlorate (KClO3) according to the following equation?2KClO3(l) + heat --> 2KCl(s) + 3O2(g)
Correct answer: C. 18 moles of oxygen
- A. 12 moles of oxygen
- B. 15 moles of oxygen
- C. 18 moles of oxygen
- D. 21 moles of oxygen
Explanation
Option C is correct. The balanced chemical equation is: 2 KClO3(l) + heat → 2 KCl(s) + 3 O2(g). This indicates that for every 2 moles of KClO3, 3 moles of O2 are produced. To find out how many moles of O2 are produced from 12 moles of KClO3, we set up the following calculation: 12 moles KClO3 * (3 moles O2 / 2 moles KClO3) = 18 moles O2. Therefore, the answer is 18 moles of oxygen. The other options are incorrect because they do not align with the stoichiometry of the reaction. Specifically, options A, B, and D propose amounts of O2 that exceed or do not reach the correct product ratio based on the given moles of KClO3.
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The s and p blocks are studied through their valence-shell configurations, periodic trends and characteristic chemical properties. Work covers the reactions of Group I and Group II elements, the behaviour of Group IV elements, and how atomic size, ionization energy, electronegativity and metallic character change across periods and down groups.
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