In the analysis of 1.5000g sample of feldspar, a mixture of the chlorides of sodium and potassium is obtained which  weighs 0.3450g. ?

Subsequently treatment with AgNO3 furnishes 0.7353 g of AgCI. What g/mol ; NaCI=58.45g/mol ; K2O=94.2g/mol; Na2O=62.00g/mol).

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  • 2 months ago
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    Weight of feldspar=1.5 gm 

    Weight of AgCl= 0.7353 gm 

    Total weight NaCl+KCl = .3450 gm 

    First the mass of AgCl will be solved obtained from NaCl and KCl 

    2NaCl=Na 2 O and 2KCl=K2O 

    Let the amount of Nacl be x gm 

    And the amount of KCl is (.3450 - x) gm 

    NaCl + AgNo3 ---> AgCl + NaNO3 

    58.5 gm NaCl gives = 143.5 gm AgCl 

    X gm = 143.5/58.5 into x  

    Similarly for KCl 

    KCl + AgNO3 ---> AgCl + KNO3 

    74.5 gm KCl gives = 143.5gm AgCl 

    (.3450 - x) gm = 143.5/74.5 x (.3450 - x ) 

    Total weight AgCl (given in the ques) = 0.7353 gm 

    ==>(143.5/58.5) into x + (143.5/74.5) x (.3450 - x)) = 0.7353  

    When we solve this we get x= 0.1332 gm 

    Amount of KCl = (0.3450-0.1332) = 0.2118gm 

    2 NaCl = Na2O 

    2x58.5 = 62 

    117gm=62gm 

    ====> 0.1332 = 62/117 x 0.1332 

    = 0.06262 gm Na2O 

    %Na2O in 1.5 gm feldspar = 0.06262/1.5 x 100 = 4.17% 

    Similarly for 2KCl = K2O 

    2x74.5 = 94 

    149 gm = 94 gm 

    149 gm KCl = 94 gm K2O 

    So ===> 0.2118 = 94/149 x 0.2118 = 0.078 gm K2O 

    % K2O in 0.5gm feldspar = 0.078/1.5 x 100 = 5.21 % K2O 

    Na2O= 4.17 % 

    K2O = 5.21% 

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