Question

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 reactions involved.

Answer 1

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

2A +2H₂O →2AOH + H₂↑

According to the question,

x+16+1 =40

x = 40 -17 

.'. x =23

[23 is the atomic weight of sodium (Na)]

Therefore, the alkali metal (A) is Na and the reaction is-

2Na + 2H₂O→ 2NaOH(aq) + H₂(g)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).

The reaction involved is -

 Al₂O₃ +2NaOH→2NaAlO₂ +H₂O