Question

. 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.
[NCERT Exemplar] 3​

Answer 1

SOLUTION:

Let the atomic weight of the alkali metal A be x.

Now,

When it reacts with water, it forms a compound B which is having molecular mass 40. (given in the question.)

So,

Let the reaction be -

$\boxed{\red{2A+2H_{2}O \longrightarrow 2A\ OH + H_{2} \uparrow}}$

Here,

2A OH is the compound B

So,

According to the question,

⇒ x + 16 + 1 = 40

⇒ x + 17 = 40

⇒ x = 40 - 17

⇒ x = 23

So,

It is the atomic weight of the Na (Sodium)

Thus,

The alkali metal A is Na(Sodium),

And the reaction is

$\boxed{\red{2Na(s)+2H_{2}O(l) \longrightarrow 2NaOH(aq) + H_{2}(g) \uparrow}}$

So,

The compound B = NaOH = Sodium Hydroxide.

The  Sodium Hydroxide reacts with aluminum oxide (Al₂O₃) to give Sodium Aluminate(NaAlO₂).

Therefore,

C is Sodium aluminate (NaAlO₂)

And the reaction which is involved -

$\boxed{\pink{Al_{2}O_{3}(s)+2NaOH(aq) \longrightarrow 2NaAlO_{2}(aq)+H_{2}O(l)}}$

Here,

NaAlO₂ = Sodium aluminate.

Al₂O₃ = Aluminum oxide

NaOH = Sodium Hydroxide.

Hence,

A = Na = Sodium

B = NaOH = Sodium hydroxide

C = NaAlO₂ = Sodium aluminate