Question

Question: Aluminium sulphate $(X)$is slightly insoluble in water. It is converted into soluble sodium sulphate by using $N{a_2}C{O_3}$in the preparation of sodium carbonate extract. Moles of $(Y)$, required for complete conversion of \[1{\rm{ }}mole\]of ()X into soluble sulphate, is :
A. 1
B. 2
C. 3
D. 4

Answer 1

Hint: We will look at the charges on the cations and anions. Aluminium and sodium cation have +3 and +1 charges respectively. Sulphate anion and carbonate anion have -2 and -2 charge respectively. In a neutral molecule, total positive charge on cations is equal to total negative charge on anions.

Complete step-by-step solution: You will get a white acceleration of aluminium hydroxide, if the two fold salt is blended in with an abundance of sodium carbonate in a fluid medium. What's more, there will likewise be advancement of carbon dioxide bubbles.
Potash alum is a hydrated twofold salt of the sulphates of potassium and aluminium \[{K_2}S{O_4}.A{l_2}{\left( {S{O_4}} \right)_3}.24{H_2}O\].
In contrast to \[{K_2}S{O_4}\], the aluminum segment, \[A{l_2}{\left( {S{O_4}} \right)_3}\], is a salt of a solid corrosive and frail base. Thus, it goes through hydrolysis in water and frees hydrogen particles.
\[A{l^{\left( {3 + } \right)}}{\rm{ }} + {\rm{ }}{H_2}O{\rm{ }} \to {\rm{ }}AlO{H^{\left( {2 + } \right)}}{\rm{ }} + {\rm{ }}{H^{\left( + \right){\;^{}}}}\]………………………………………………………………… (I)
Sodium carbonate being a salt of a feeble corrosive and solid base produces hydroxide particles by hydrolysis.
\[CO_3^{\left( {2 - } \right)}{\rm{ }} + {\rm{ }}{H_2}O{\rm{ }} \to {\rm{ }}HCO_3^{( - )}{\rm{ }} + {\rm{ }}O{H^{\left( - \right)}}\]…………………………………………………………………. (II)
The ${H^{( + )}}$ions from (I) and \[HCO_3^{( - )}\]ions from (II) neutralise each other and liberates carbon dioxide gas.
\[HCO_3^{\left( - \right)}{\rm{ }} + {\rm{ }}{H^{\left( + \right)}}{\rm{ }} = {\rm{ }}C{O_2}{\rm{ }} + {\rm{ }}{H_2}O\]
The \[AlO{H^{\left( {2 + } \right)}}\]particles present in the arrangement will join with the \[O{H^{\left( { - {\rm{ }}} \right)}}\]particles to shape insoluble aluminum hydroxide, \[Al{\left( {OH} \right)_3}\], as a white hasten.

The rest of the arrangement will contain sodium sulfate \[\left( {N{a_2}S{O_4}} \right),\]another significant result of the response, alongside the unreacted potassium sulfate.
When aluminium sulphate reacts with sodium carbonate, it is unbalanced. Hence, it should be balanced. Hence when one moles of aluminium sulphate reacts with three moles of the sodium carbonate.
\[1{\rm{ }}mole\] of $(X)$reacts with \[{\rm{3 }}moles\] of $(Y)$to form soluble sulphate which is a double salt.
$A{l_2}{(S{O_4})_3} + 3N{a_2}C{O_3} + x{H_2}O \to A{l_2}{(S{O_4})_3} \cdot 3N{a_2}C{O_3} \cdot x{H_2}O$
This is the balanced equation.

Note:When aluminium sulphate reacts with sodium carbonate, it is unbalanced. Hence, it should be balanced. Hence when one moles of aluminium sulphate reacts with three moles of the sodium carbonate.