3. Consider the reaction (where H:NCH,CH2NH2 is ethylenediamine): Fe3. (aq) + 3

Question

3. Consider the reaction (where H:NCH,CH2NH2 is ethylenediamine): Fe3. (aq) + 3 (H2NCH2CHANH) (aq) [Fe(H2NCHCH,NH2hP (aq) Calculate the number of moles of iron/(lI) Ion when you obtain 8.0 ml of 0.40 g/mL of FeChy 6H20 solution. (b) Calculate the number of moles of ethylenediamine in 4.0g of H,NCH,CH,NHs c) Since iron(ill) and ethylenediamine react in a 1:3 mole ratio to form [Fe(H NCH,CH/NHahl (aq), what is the limiting reactant in a mixture of 8.0 ml of 0.40g/mL of FeCly 6H 0 and 4.0 g of ethylenediamine. (d) How many moles of [Fe(H2NCHCHNOhproduct do you expect to form, in theory? in Part B.4 of the Experimental Procedure, you are requested to store the green crystals of the synthesized coordination compound in the absence of light in a closed drawer. What is the purpose of storing the green crystals in the dark? 4.

Explanation / Answer

Fe+3 + 3 (2HNCH2CH2NH2) (aq) ------> [Fe(2HNCH2CH2NH2)]^+3 (aq)

volume of the compound = 8 mL

density of the solution = 0.4 g/mL

mass of the solution = Density * volume of the solution

                      = 0.4 g/mL * 8 mL = 3.2 g

molar mass of the compound FeCl3.6H2O = 270.3 g/mol

number of moles of FeCl3.6H2O = 3.2/270.3 = 0.01184 moles

1 mole of FeCl3.6H20 produces 1 mole of Fe+3 ion hence,

number of moles of Fe+3 ion = 0.01184 moles

number of moles of ethylene diamine = 4g /60.27 g/mol = 0.06637 moles

From the reaction 1 mole of Fe+3 require 3 moles of Ethyleneaiamine, so

number of moles of ethylenediamine required in the above reaction = 3 * 0.01184 moles

number of moles of ethylenediamine = 0.03552 moles

hence limiting reagent is Fe+3 ion and excess reagent is ethylenediamine

From the reaction 1 mole of Fe+3 produces 1 mole of the product.

0.01184 moles of Fe+3 produces 0.01184 moles of the product.

amount of the product produces = 0.01184 moles.

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