What is the theoretical yield of [Ni(en)3]Cl2 ?

Consider the reaction:

NiCl2·6H2O + 3en → [Ni(en)3]Cl2 + 6H2O

You start with 2.19 mol of NiCl2·6H2O and 5.80 mol of en (en = H2NCH2CH2NH2).

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  • 3 weeks ago

    5.80 moles of en would react completely with 5.80 x (1/3) = 1.93 moles of NiCl2·6H2O, but there is more NiCl2·6H2O  present than that, so NiCl2·6H2O is in excess and en is the limiting reactant.

    (5.80 mol en) x (1 mol [Ni(en)3]Cl2 / 3 mol en) = 1.93 mol [Ni(en)3]Cl2

    (5.80 / 3) mol [Ni(en)3]Cl2 x (309.8944 g [Ni(en)3]Cl2 / mol) =

    599 g [Ni(en)3]Cl2 in theory

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