How do you calculate the weight of ethylenediamine (en) used? (note: 30% en solution (15mL) was neutralised with 5M HCl (2.5mL) density of en also given to aid in calculation :0.90g/mL)

I’m a little confused by the data given and here’s why.
Ethylenediamine (en) or ##(CH_2)_2(NH_2)_2## is a weak which implies that you’ve essentially performed a weak base – strong acid .
Because en is dibasic you’ll have 2 equivalence points on the titration curve and the after will be less than 7.
Now if you’ve used 15 mL of a 30% en solution in the titration the 2.5 mL volume of ##HCl## solution used is far too little.
A ##30% v/v## solution would have 30 mL of en in every 100 mL of solution; if you’ve used 15 mL of solution then the volume of en is
##30% = V_(en)/V_(solution) * 100 => V_(en) = (30 * V_(solution))/100##
##V_(en) = (30 * 15 mL)/100 = 4.5 mL##
Using its given will give you the mass
##4.5cancel(mL) * 0.90 g/(1cancel(mL)) = 4.05 g##
Here’s what doesn’t seem right to me. In this case the number of moles of en would be
##4.05cancel(g) * 1 mole/(60.10cancel(g)) = 0.0674 moles en##
The first equivalence point would have the number of moles of en equal to the number of moles of HCl which is
##C = n/V => n = C * V##
##n_(HCl) = 5 M * 2.5 * 10^(-3)L = 0.0125 moles HCl##
In this case It’sclear that you’ve used too little ##HCl## and is a long way ahead i.e. you need more ##HCl##.
Another approach I’d take is to work backwards from the number of moles of ##HCl## to get the number of moles of en.
The second equivalence point requires 2 times more moles of ##HCl## than of en which means that
##n_(en) = 0.0125 moles/2 = 0.00625 moles en##
This is equivalent to
##0.00625cancel(moles en) * 60.10 g/(1 cancel(mole en)) = 0.376 g en ## or
##rho = m/V => V_(en) = m/(rho) = (0.376cancel(g))/(0.90cancel(g)/mL) = 0.42 mL##
which corresponds to
##V_(solution) = (V_(en) * 100)/30 = 1.4 mL en solution##
As I can see it you’ve either used 1.5 mL of solution instead of 15 mL or 25 mL of HCl instead of 2.5 mL.