According to the question,
- 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
2Na + 2H2O → 2NaOH + H2
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
- A is sodium
- B is Sodium Hydroxide
- C is Sodium Aluminate