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Question 64
An element A burns with golden flame in 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

1. When sodium is exposed to air it burns out with a golden flame. Hence, A is Na.

2. The element having atomic number 17 is chlorine. Hence B is Cl2.

3. When sodium reacts with chlorine, it forms sodium chloride (NaCl) which is an ionic compound. The reaction involved is:

2Na(s)+Cl2(g)2NaCl(s)

Hence, C is NaCl.

4. On electrolysis, NaCl solution forms a sodium compound i.e. sodium hydroxide and releases chlorine and hydrogen gas. The reaction involved is:

2NaCl (aq)+2H2O (l)2NaOH (aq)+Cl2 (g)+H2(g)

Hence, D is NaOH.


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