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. 

Explanation

Let the atomic weight of alkali metal A is x.
When it reacts with water it forms a compound B having molecular mass 40.
Let, the reaction is
`2A + 2H_(2) O rarr 2AOH + H_(2)`
According to question, molecular mass of compound B = 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 + 2H_(2)O rarr 2NaOH + H_(2)`
So, compound B is sodium hydroxide (NaOH). Sodium hydroxide reacts with aluminium oxide `(Al_(2)O_(3))` to give sodium aluminate. Thus, C is sodium aluminate. The reaction involved is
`Al_(2)O_(3) + 2NaOH rarr 2NaAlO_(2) + H_(2)O`