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  • Solve! - s- Block Elements ( Alkali and Alkaline earth metals ) - NEET

The compound A on heating gives a colourless gas and a residue that is dissolved in water to obtain B. Excess of CO_{2} is bubbled through aqueous solution of B, C is formed which is recovered in the solid form. Solid C on gentle heating gives back A. The compound is

  • Option 1)

    C_{a}SO_{4}. 2H_{2}O

  • Option 2)

    C_{a}CO_{3}

  • Option 3)

    Na_{2}CO_{3}

  • Option 4)

    K_{2}CO_{3}

 

Answers (1)

As we learned in concept no

Calcium oxide/quick lime/time -

Its aqueous suspension is known as slaked lime, Ca(OH)_{2}

- wherein

CaO+H{_2}O\xrightarrow{hissing sound} Ca(OH)_{2}+heat

 

 

Action heat on CaCO3 -

Decomposes to evolve carbon di oxide

- wherein

CaCO_{3}\overset{1200K}{\rightarrow}CaO+CO_{2}

 

 

Action of CO2 on limewater -

Turns milky due to the formation of calcium carbonate

- wherein

Ca(OH)_{2}+CO_{2}\rightarrow CaCO_{3}+H_{2}O

 

 CaCO_{3}(s)\overset{\Delta}{\rightarrow}CO_{2}(g)9+CaO(s)

CaO(s)+H_{2}O\longrightarrow Ca(OH)_{2}

Ca(OH)_{2}+2CO_{2}+H_{2}O\longrightarrow Ca(HCO_{3})_{2}

 


Option 1)

C_{a}SO_{4}. 2H_{2}O

This is incorrect  option

Option 2)

C_{a}CO_{3}

This is correct  option

Option 3)

Na_{2}CO_{3}

This is incorrect  option

Option 4)

K_{2}CO_{3}

This is incorrect  option

Posted by

subam

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