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 aluminum oxide. Identify A, B, and C and give the reaction involved.

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Solution

Step 1: Alkali metals:
Alkali metals are those metals that have their last electron present in s subshell as $n s^{1}$ .
Alkali metals in reaction with water produce hydroxide ions.
Step 2: Determination of Alkali metal:
When alkali metal reacts with water it produces hydroxide of molecular weight $40$ . Then, $M + O + H = 40$ .
Now putting the actual atomic weights of the atoms we get $M + 16 + 1 = 40$ .
Thus the calculated atomic weight of the alkali metal is $M = 40 - 17 = 23$ .
Thus, Alkali metal A is Sodium that has $11$ atomic number.
Step 3: Reaction of Sodium with water:
When Sodium reacts with water, it forms Sodium hydroxide $(NaOH)$ along with the liberation of hydrogen $(H_{2})$ gas.
The chemical reaction can be depicted as:
$2 Na (s) + 2 H_{2} O (l) \to 2 NaOH (aq) + H_{2} (g)$
Thus, compound B is Sodium hydroxide $(NaOH)$ .
Step 4: Reaction of Sodium hydroxide with Aluminum oxide:
When Sodium hydroxide reacts with Aluminium oxide, it forms Sodium aluminate $({NaAlO}_{2})$ and water $(H_{2} O)$ .
The chemical reaction is depicted as:
${Al}_{2} O_{3} (s) + 2 NaOH (aq) \to 2 {NaAlO}_{2} (s) + H_{2} O (l)$
Thus, compound C is Sodium aluminate $({NaAlO}_{2})$ .
Thus, compounds A, B, and C are alkali metals, Sodium hydroxide, and Sodium aluminate respectively.

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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 aluminum oxide. Identify A, B, and C and give the reaction involved.

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 aluminum oxide. Identify A, B, and C and give the reaction involved.