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 sodium because molar mass of NaOH is 40; which can be shown as follows:
Na (23) +O (16) +H(1)=40
So, B is NaOH
When sodium hydroxide is treated with aluminium oxide, we get sodium aluminate. So, C is sodium aluminate. Following reaction happens in this case:
Al2O3 + 2NaOH → 2NaAlO2 + H2O
