Question

A metal $M$ reacts with sodium hydroxide to give a white precipitate $X$ which is soluble in excess of $NaOH$ to give $Y$. Compound $X$ is soluble in $HCl$ to form a compound $Z$. Identify $M,X,Y$ and $Z$.

• A. $Si-M,$ $Si{O}_2-X$, $N{a}_{2}Si{O}_3-Y$, $SiC{l}_4-Z$
• B. $Al-M,$ $Al(OH{)}_3-X$, $NaAl{O}_2-Y$, $AlC{l}_3-Z$
• C. $Mg-M,$ $Mg(OH{)}_3-X$, $NaMg{O}_3-Y$, $MgC{l}_2-Z$
• D. $Ca-M,$ $Ca(OH{)}_2-X$, $N{a}_{2}C{O}_3-Y$, $NaHC{O}_3-Z$

Answer 1

The correct option is B $Al-M,$ $Al(OH{)}_3-X$, $NaAl{O}_2-Y$, $AlC{l}_3-Z$

$\because$ Metal $M$ on reaction with $NaOH$ gives white precipitate which is soluble in $NaOH$.

$\therefore$ Metal $M$ is Aluminium.

$\underset{\text{Metal M}}{Al}+3NaOH⟶\underset{\text{white ppt.}}{Al{\left(OH\right)}_3}+3{Na}^{+}$

Thus, $X$ is $Al{\left(OH\right)}_3$

$Al{\left(OH\right)}_3+\underset{Excess}{NaOH}⟶NaAl{O}_2+2H_{2}O$

Thus, $Y$ is $NaAl{O}_2$.

$\because$ compouns X when soluble in $HCl$ forms compound Z.

$\therefore Al{\left(OH\right)}_3+3HCl⟶Al{Cl}_3+3H_{2}O$

Thus, $Z$ is $Al{Cl}_3$.

There is no option which shows $Y$ as $NaAl{O}_2$.