An element A burns with golden flame in the air. It reacts with another element B, atomic number 17 to give a product C. An aqueous solution of product C on electrolysis gives a compound D and liberates hydrogen. Identify A, B, C and D. Also write down the equations for the reactions involved.

Answer:

Element A is Sodium because it will burn with a golden flame in the air.

Element B is Chlorine for its atomic number is 17.

Product C is Sodium Chloride. The reaction can be expressed as

2Na +Cl2 → 2NaCl

On electrolysis of aqueous NaCl solution, the compound D, sodium hydroxide(NaOH), is formed. The equation can be written as

2NaCl + 2H2O → 2NaOH + Cl2 + H2

Hence

  • A is Soadium (Na)
  • B is Chlorine(Cl)
  • C is Sodium chloride (NaCl)
  • D is Sodium hydroxide (NaOH)

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