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4.0 g of ferrous ammmonium sulphate, FeSO4(NH4)2SO4 6H2O, is used. Since the oxalate is in excess, calculate the theoretical yield of the iron complex.

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  • Biology -

    Check some of the following links to see if anything there can help you:

    http://search.yahoo.com/search?fr=mcafee&p=4.0+g+of+ferrous+ammmonium+sulphate%2C+FeSO4%28NH4%292SO4+6H2O%2C+is+used.+Since+the+oxalate+is+in+excess%2C+calculate+the+theoretical+yield+of+the+iron+complex

    Sra

  • Biology -

    Calculate the theoretical yield for K3[Fe(C2O4)3]*3H2O; 491.258 g/mol

    Mass of Ferrous Ammonium Sulfate Hexahyrdreate: 4.01 g, 392.17 g/mol

    [1] FeSO4∙(NH4)2SO4∙ 6H2O + H2C2O4∙ 2H2O ---> FeC2O4 + (NH4)2SO4 + H2SO4 + 8H2O

    [2] 6 FeC2O4 + 3H2O2 + 6K2C2O4∙ H2O --->
    4K3[Fe(C2O4)3]∙ 3 H2O + 2 Fe(OH)3 + 6 H2O

    [3] 2 Fe(OH)3 + 3 H2C2O4∙ 2H2O + 3 K2C2O4∙ H2O ----> 2 K3[Fe(C2O4)3]∙ 3H2O + 9 H2O


    4.01g * 1mol/392.17g = 0.01 mol FeSO4∙(NH4)2SO4∙ 6H2O

    0.01 mol * 1/1 = 0.01 mol FeC2O4

    0.01 mol * 4/6 = 0.00667 mol K3[Fe(C2O4)3]*3H2O

    0.00667 mol * 491.258g/1mol = 3.28 g K3[Fe(C2O4)3]*3H2O

    0.01 mol * 2/6 = 0.0033 mol Fe(OH)3

    0.0033 * 2/2 = 0.0033 mol K3[Fe(C2O4)3]*3H2O

    0.0033 mol * 491.258g/1mol = 1.62 g K3[Fe(C2O4)3]*3H2O

    3.28 + 1.62 = 4.9 g K3[Fe(C2O4)3]*3H2O


    The steps are all correct over all answer is wrong however it should be 5.011 rounded off. Make sure you plugin the correct number following the procedure.

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