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3 September, 10:47

Nick and Kel reacted 2.80 g of NiCl2· 6 H2O with 7.00 mL of 4.00 M C2H8N2. They recovered 2.40 g of product, [Ni (C2H8N2) 3] Cl2. At the end, the filtrate was a dark blue solution. The litmus test on the filtrate solution left the red litmus paper unchanged in color.

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  1. 3 September, 12:09
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    Actual Product yield = 0.028 x 308.7

    Explanation:

    Limiting reactant

    moles of NiCl2.6H2O = 2.80 g/237.7 g/mol = 0.011 mol

    moles of C2H8N2 = 4 M x 7.00 ml = 0.028 mol

    for formation of complex,

    If all of NiCl2.6H2O is consumed we would need = 0.011 x 3 = 0.033 mols of C2H8N2

    If all of C2H8N2 is consumed we reuire = 0.028/3 = 0.009 mols ofNiCl2.6H2O

    since moles of NiCl2.6H2O available is in excess, C2H8N2 is limiting reactant

    Actual Product yield = 0.028 x 308.7
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