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Related Topic: Chlorine and its Compounds

Related Subtopic: Laboratory preparation of chlorine



Questions 6.2

  1. State the uses of water and concentrated sulphuric (VI) acid in the set-up.
  2. Identify the four products in Flask 1.
  3. Write a complete formula equation for the reaction in Flask 1.
  4. Explain why this reaction is considered as a reduction-oxidation (redox) reaction.
  5. Complete the following equations for alternative reactions that also produce chlorine.
    MnO2(s) + HCl(aq)
    PbO2(s) + HCl(aq)


Answers to Questions 6.2


  1. Water: It absorbs and therefore removes hydrogen chloride (HCl) gas released by concentrated hydrochloric acid. Concentrated sulphuric (VI) acid dries the gas by absorbing water vapour.
  2. Manganese (II) chloride (MnCl2)
    Potassium chloride (KCl)
    Water (H2O)
    Chlorine (Cl2)

  3. 2KMnO4(s) + 16HCl(aq) → 2MnCl2(aq) + 2KCl(aq) + 8H2O(l) + 3Cl2(g)
  4. Hydrogen is removed from its compound with chlorine (HCl); so chlorine is oxidized. Oxidation state of manganese reduces from +7 to +2; so manganese is reduced.
  5. MnO2(s) + 4HCl(aq) MnCl2(aq) + 2H2O(l) + Cl2(g)
    PbO2(s) + 4HCl(aq) PbCl2(aq) + 2H2O(l) + Cl2(g)


    6. NB: Balancing long equations such as in Question 3 can be hectic. However, if you can recall one critical number, such as 16 on HCl, the rest will follow.




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