Na_2CO_3 is prepared by the Solvay process,bu | vedclass

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(C) In the Solvay process,$NaHCO_3$ is precipitated because it is sparingly soluble in water and can be easily separated by filtration. However,in the

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$KHCO_3$ is appreciably soluble

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(C) In the Solvay process,$NaHCO_3$ is precipitated because it is sparingly soluble in water and can be easily separated by filtration. However,in the case of potassium,the reaction is: $(NH_4)HCO_3 + KCl \longrightarrow KHCO_3(aq) + NH_4Cl(aq)$. Since $KHCO_3$ is appreciably soluble in water,it does not precipitate out of the solution. Therefore,it cannot be isolated from the reaction medium to be converted into $K_2CO_3$ by heating.

Metals	Flame color
$1$. $Li$	$a$. Crimson red
$2$. $K$	$b$. Violet
$3$. $Rb$	$c$. Red-violet
$4$. $Cs$	$d$. Blue

$Na_2CO_3$ is prepared by the Solvay process,but $K_2CO_3$ cannot be prepared by the same process because:

Similar Questions

Match the following:

Metals Flame color

$1$. $Li$ $a$. Crimson red

$2$. $K$ $b$. Violet

$3$. $Rb$ $c$. Red-violet

$4$. $Cs$ $d$. Blue

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