Question

An alkali metal A Gives a compound B(molecular mass=40)in 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
.

Answer 1

Answer:

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 +H2

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+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+

Al2O3+2NaOH------>2NaAlO2+H2O

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