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)Otounderset((B))(2AOH)+H_(2)uparrow`
According to question, molecular mass of compound
`B=x+16+1=40`
`x=40-17=23`
it is the atomic weight of Na (sodium), Therefore, the alkali metal (A) is Na and the reaction is
`underset((A))(2Na)+2H_(2)Otounderset((B))(2NaOH)(aq)+H_(2)(g)uparrow`
compound B is sodium hydroxide (NaOH)
Sodium hydroxide reacts with aluminium oxide `(Al_(2)O_(3))` to give sodium aiuminate `(NaAIO_(2))`
Thus, C is sodium aluminate `NaAIO_(2)`
The reaction invoved is `Al_(2)O_(3)+underset((B))(2NaOH)tounderset("Sodium aluminate(C)")(2NaAlO_(2))+H_(2)O`