Question

A compound (X) imparts a golden yellow flame and shows the following reactions :
(i) Zinc powder, when boiled with the concentrated aqueous solution of (X) dissolves and hydrogen, is evolved.
(ii) When an aqueous solution of (X) is added to an aqueous solution of stannous chloride, a white precipitate is obtained first which dissolves in excess of solution of (X)
Identify (X) .

• A. $NaOH$
• B. $KOH$
• C. $Mg(OH{)}_2$
• D. None of these

Answer 1

The correct option is A $NaOH$

(a) The compound (X) is sodium hydroxide (NaOH). Because Na compounds imparts golden-yellow flame.

(i) When zinc is boiled with NaOH, soluble sodium zincate is obtained along with evolution of hydrogen gas.
$Zn+2NaOH⟶N{a}_{2}Zn{O}_2+H_2$

(ii) Stannous chloride reacts with NaOH to form white precipitate of tin(II) hydroxide which dissolves in excess NaOH to form $N{a}_{2}Sn{O}_2$

$SnC{l}_2+2NaOH⟶Sn\underset{white}{(OH{)}_2}\downarrow +2NaCl$

$Sn(OH{)}_2+2NaOH⟶\underset{soluble}{N{a}_{2}Sn{O}_2}+2H_{2}O$

Option A is correct.