Question

1. How many moles of ethylenediamine (H₂NCH₂CH₂NH₂) are contained in 10.0 mL of 25.0% ethylenediamine solution? The density of the solution is 0.950 g/mL   

2. Under the conditions employed in this experiment, 10.0 mL of 25.0% ethylenediamine were reacted with 1.80 g of NiCl₂X6H₂O. What is the theoretical yield of [Ni(en)₃]Cl₂? The density of the ethylenediamine solution is 0.950 g/mL.

I greatly appreciate your help. Full steps would be lovely. My tired glucose-starved brain cannot process this.

Answer 1

1. 25% ethylenediamine solution, means 25 g ethylenediamine present in 100 g solution .

    so that, volume of solution = m/d = 100/0.95 = 105.26 ml

   10 ml solution containing ethylenediamine = 10*25/105.26 = 2.375 g

    no of mole of ethylenediamine = 2.375/60.2172 = 0.0394 mol

2)

       no of mole of NiCl2*6H2O = 1.80/237.69108 = 0.0076 mol

   NiCl2*6H2O + 3 ethylenediamine ----> [Ni(en)3]Cl2 + 6H2O

   from formula , 3 mole ethylenediamine = 1 mol NiCl2*6H2O

   limiting reactant = NiCl2*6H2O

    theoretical yield of [Ni(en)3]Cl2 = 0.0076*1/1*350.212 = 2.66 g