Q54 of 65 Page 24

An alkali metal A gives a compound B (molecular mass = 40) on reacting with water. The compound B gives a soluble compound C on treatment with aluminium oxide. Identify A, B and C and give the reaction involved.


A is Na, B is NaOH, C is NaAlO2
2Na (s) + 2H2O → NaOH+ H2 (g)
Al2O3 + is NaOH → 2NaAlO2 + H2O


Let ‘s suppose the atomic weight of alkali metal A be x. When it reacts with water, it forms a compound B having molecular mass 40.


Therefore, x + 16 + 1 = 40
x = 40 –17 = 23
The atomic weight of Na (sodium) is 23.
Therefore, the alkali metal (A) is Na. Sodium reacts with water to form sodium hydroxide. So, compound B is sodium hydroxide (NaOH).
Sodium hydroxide reacts with aluminium oxide (Al2O3) to give sodium aluminate (NaAlO2). Thus, C is sodium aluminate (NaAlO2).


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