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Question

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.


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Solution

Element A:

  • The element A is Sodium (Na) because it will burn with a golden flame in the air.
  • Which belongs to alkaline metals.
  • Which is a reactive metal.

Element B:

  • The element B is Chlorine (Cl) and its atomic number is 17.
  • Which is an electronegative element.
  • It belongs to halogen family.

Product C :

  • The product C is Sodium Chloride.
  • It is a salt obtained by the neutralization reaction.
  • The reaction of sodium with chlorine gas gives sodium chloride, the equation can be expressed as

2Na(s)+Cl2(g)2NaCl(s)(Sodium)(Chlorinegas)(Sodiumchloride)

Compound D:

  • On electrolysis of aqueousNaCl solution, the compound D, sodium hydroxide(NaOH) and hydrogen gas is formed.
  • The equation can be written as

2NaCl(aq)+2H2O(l)2NaOH(aq)+Cl2(g)+H2(g)(Sodiumchloride)(Water)(Sodiumhydroxide)(Chlorine)(Hydrogengas)

Hence

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

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