Question

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

Answer 1

Alkali metals are sodium,potassium,lithium etc.

On reacting with water , they give metal hydroxide M_k (OH)_n

Hydroxide molecular mass = 17..
As molecular mass of the metal hydroxide is 40, n = 1 or 2 and  correspondingly the molecular mass of metal is 23 or 6.
That is metal A is  either sodium or lithium.

Compound B is sodium hydroxide or lithium hydroxide.Lithium hydroxide-molecular mass is not 40.

So metal A is Sodium:Na compound B is sodium hydroxide.
aluminum oxide:   Al₂ O₃ + 2 Na OH  +3 H₂O ==> 2 Na Al (OH)₄
          
Compound C :colourless solution in water of  compound sodium tetra aluminate.

Answer 2

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 + 2H20 = 2AOH + H2 (G)

According to the question,

x + 16 + 1 = 40 (Given)

x = 40-17 = 23

It is the atomic weight of Na (sodium).  

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

2Na + 2H2O = 2NaOH + 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+

Al2O3 + 2NaOH = 2NaAlO2 + H20