A metal (X) dissolves in both dilute HCl and | vedclass

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(B) $Al$ $(X)$ is an amphoteric metal that dissolves in both dilute $HCl$ and dilute $NaOH$ to liberate $H_2$ gas.$2Al + 6HCl \longrightarrow 2AlCl_3

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$Al$ and $Al(OH)_3$

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(B) $Al$ $(X)$ is an amphoteric metal that dissolves in both dilute $HCl$ and dilute $NaOH$ to liberate $H_2$ gas.$2Al + 6HCl \longrightarrow 2AlCl_3 + 3H_2$$2Al + 2NaOH + 2H_2O \longrightarrow 2NaAlO_2 + 3H_2$When $NH_4Cl$ and excess $NH_4OH$ are added to the $AlCl_3$ solution,$Al(OH)_3$ $(Y)$ is precipitated.$AlCl_3 + 3NH_4OH \longrightarrow Al(OH)_3 \downarrow + 3NH_4Cl$Similarly,adding $NH_4Cl$ to the $NaAlO_2$ solution (formed by $Al$ in $NaOH$) also precipitates $Al(OH)_3$ $(Y)$ due to the common ion effect and hydrolysis of the aluminate ion.$NaAlO_2 + NH_4Cl + H_2O \longrightarrow Al(OH)_3 \downarrow + NaCl + NH_3$Thus,$(X)$ is $Al$ and $(Y)$ is $Al(OH)_3$.

List-$I$ (Family)	List-$II$ (Symbol of Element)
$A$. Pnictogen (group $15$)	$I$. $Ts$
$B$. Chalcogen	$II$. $Og$
$C$. Halogen	$III$. $Lv$
$D$. Noble gas	$IV$. $Mc$

Similar Questions

Of the following,the non-existent compound is

Match the List-$I$ with List-$II$.
List-$I$ (Family) List-$II$ (Symbol of Element)
$A$. Pnictogen (group $15$) $I$. $Ts$
$B$. Chalcogen $II$. $Og$
$C$. Halogen $III$. $Lv$
$D$. Noble gas $IV$. $Mc$

Choose the correct answer from the options given below:

In the following reaction,identify the nature of the product formed by the complete hydrolysis of the underlined atom at $R.T.$:
$\underline{N}_2O_5 + H_2O \longrightarrow 2HNO_3$

Which one of the following is correct with respect to basic character?

$HCl$ reacts with $NH_3$ to produce white fumes of which compound?

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