Q1: You measure out 1.9 mL of an aqueous ethylenediamine (H 2 NCH 2 CH 2 NH 2 )

Question

Q1: You measure out 1.9 mL of an aqueous ethylenediamine (H2NCH2CH2NH2) solution. This aqueous solution has a density of 0.950 g/mL and is 25.0% by mass ethylenediamine. How many moles of ethylenediamine are in the 1.9 mL?

Q2:  Consider the reaction: NiCl2·6H2O + 3en [Ni(en)3]Cl2 + 6H2O You start with 0.63 mol of NiCl2·6H2O and 1.67 mol of en (en = H2NCH2CH2NH2). What is the theoretical yield of [Ni(en)3]Cl2 ?

Q3: In a titration experiment, 9.3 mL of an aqueous HCl solution was titrated with 0.4 M NaOH solution. The equivalence point in the titration was reached when 9.5 mL of the NaOH solution was added. What is the molarity of the HCl solution?

Explanation / Answer

Q1: You measure out 1.9 mL of an aqueous ethylenediamine (H2NCH2CH2NH2) solution. This aqueous solution

has a density of 0.950 g/mL and is 25.0% by mass ethylenediamine. How many moles of ethylenediamine are in

the 1.9 mL?
Mass of solution = density*volume=0.95*1.9= 1.805 g
100 gm of solution has 25 g of ethylenediamine
1.805 g of solution 25/100*1.805 = 0.475 g of ethlenediamine
Molar mass of ethylenediamine, MW = 60.1 g/mol
Moles of ethylenediamine, n = 0.475/60.1 = 0.008 mol

Q2: Consider the reaction: NiCl2·6H2O + 3en [Ni(en)3]Cl2 + 6H2O
You start with 0.63 mol of NiCl2·6H2O and 1.67 mol of en (en = H2NCH2CH2NH2).
What is the theoretical yield of [Ni(en)3]Cl2 ?

1 mol of NiCl2·6H2O requires 3 moles of en
0.63 moles of NiCl2·6H2O require 3*0.63 = 1.89 moles of en
Given 1.67 moles of en, so en is the limiting reagent.

3 moles of en produces 1 mol of [Ni(en)3]Cl2
1 moles of en produces 1/3 mol of [Ni(en)3]Cl2
1.67 moles of en produces 1.67/3= 0.557 moles of [Ni(en)3]Cl2
Molar mass of [Ni(en)3]Cl2 = 309.9 g/mol
Theoretical yield of [Ni(en)3]Cl2 = 0.557 mol = 0.557*309.9= 172.511 g

Q3: In a titration experiment, 9.3 mL of an aqueous HCl solution was titrated with 0.4 M NaOH solution.

The equivalence point in the titration was reached when 9.5 mL of the NaOH solution was added.
What is the molarity of the HCl solution?
milliequivalent of 9.5ml of 0.4 M NaOH = 9.5*0.4 = 3.8
Let molarity of HCl = X M
milliequivalent of 9.3ml of X M HCl = X*9.3
At equivalence point, milliequivalent of HCl = milliequivalent of NaOH
X*9.3=3.8
X = 3.8/9.3= 0.41 M

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