Question

Which of the following statement is incorrect?

• A. $Al$ reacts with excess $NaOH$ to give $Al{\left(OH\right)}_3$
• B. $NaH{CO}_3$ on heating gives ${Na}_2{CO}_3$
• C. Pure sodium metal dissolves in liquid ammonia to give blue solution
• D. $NaOH$ reacts with glass to give sodium silicate

Answer 1

Answer: (A)

$Al$ reacts with excess $NaOH$ to give $Al{\left(OH\right)}_3$

Sodium aluminate is also formed by the action of sodium hydroxide on elemental aluminium which is an amphoteric metal. The reaction is highly exothermic once established and is accompanied by the rapid evolution of hydrogen gas. The reaction is written as:
$2Al+2NaOH+{2H}_{2}O⟶2NaAl{O}_2+3H$
$2NaHC{O}_{3\left(S\right)}⟶N{a}_{2}C{O}_{3\left(S\right)}+C{O}_{2\left(g\right)}+H_2{O}_{\left(l\right)}$ This statment is correct.

By the reaction of sodium with liquid ammonia sodium ammide can be prepared.Sodium (alkali metal)dissolves in liquid ammoia solution to form a blue solution characteristic of a solvated electron in liquid ammonia, which is conducting in nature, due to ammoniated electrons (solvated electron).
This statment is correct.

Sodium silicate liquid can also be prepared directly by dissolving silica sand under pressure in a heated aqueous solution of caustic soda (sodium hydroxide, NaOH):
$2NaOH+{SiO}_2⟶N{a}_{2}O.{SiO}_2+H_{2}O$
This statment is correct.