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The products obtained during treatment of hard water using Clark’s method are:

Option 1 -

CaCO3 and MgCO3

Option 2 -

Ca(OH)2 and Mg(OH)2

Option 3 -

CaCO3 and Mg(OH)2       

Option 4 -

Ca(OH)2 and MgCO3

0 2 Views | Posted 2 months ago
Asked by Shiksha User

  • 1 Answer

  • V

    Answered by

    Vishal Baghel | Contributor-Level 10

    2 months ago
    Correct Option - 3


    Detailed Solution:

    In Clark’s method lime water is used

    Ca (HCO3)2+2Ca (OH)2→2CaCO3+2H2OMg (HCO3)2+2Ca (OH)2→2CaCO3+Mg (OH)2+2H2O

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Reaction of BeCl2 with LiAlH4 gives:

(A) AlCl3

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