Question

An alkali metal a gives a compound b molecular mass = 40 on reacting with water the b gives a soluble compound seen on treatment with aluminium oxide identify a b and c

Answer 1

Answer:

A=Na=sodium

B=NaOH

As 23+16+1=40

C=sodium aluminate

2NaOH+Al2O3=2NaAlO2+H20

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Answer 2

Answer:

Here's Your Answer

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Let the atomic weight of alkali metal A be x. When it reacts with water, it forms a compound B having molecular mass 40. Let the reaction be

2A + 2H2O ------->2AOH + H2O

(B)

According to the question,

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 + 2H2O -----------> 2NaOH (aq) + H2(g)

(A) (B)

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

(B) (C)