Placement of an LED in a circuit When an LED is connected in a circuit it needs a resistor so that excess current does not pass through it, but when a resistor is placed after the LED, it works the same as when the resistor is placed before the LED. Why? It has a high amount of current flowing through it as the resistor opposes the current after it flows through the LED?
 A: 
Ok maybe that part of Kirchoffs law confuses me alot.. where is it
  applicable in my question

Kirchhoff's circuit laws are fundamental to (lumped element) circuit theory.
In the case of Kirchhoff's current law (KCL), which is essentially that the sum of currents entering a node (junction of two or more circuit elements) equals the sum of currents leaving the node, we learn that series connected circuit elements have identical current through.
That is, according to KCL, all of the current out of the LED is in to the resistor.  If you were to swap their order, then all of the current out of the resistor is in to the LED.  In either case, both the LED and resistor have identical current through.
What then does the resistor do? Since the voltage across the (forward biased) LED is essentially constant over a wide range of current through, the voltage across the resistor is essentially fixed.  Thus, by Ohm's law, the series current is determined by the value of the resistance.

From a comment:

how do the electrons know that there is resitance ahead 'go slow'?

I directly quote from Griffith's "Introduction to Electrodynamics, 4th Edition", section 7.1.2, page 303:

If you think about a typical electric circuit - a battery hooked up to
  a light bulb, say (Fig 7.7) - a perplexing question arises:  In
  practice, the current is the same all the way around the loop; why
  is this the case, when the only obvious driving force is inside the
  battery?  Off hand, you might expect a large current in the battery
  and none at all in the lamp.  Who's doing the pushing, in the rest of
  the circuit, and how does it happen that this push is exactly right to
  produce the same current in each segment?  What's more, given that the
  charges in a typical wire move (literally) at a a snail's pace, (see
  Prob. 5.20), why doesn't it take half an hour for the current to reach
  the light bulb?  How do all the charges know to start moving at the
  same instant?
Answer:  If the current were not the same all the way around (for instance, during the first split second after the switch is closed),
  then charge would be piling up somewhere, and - here's the crucial
  point - the electric field of this accumulating charge is in such a
  directions as to even out the flow.  Suppose, for instance, that the
  current into the bend in Fig. 7.8 is greater than the current out.
  Then charge piles up at the "knee", and this produces a field aiming
  away from the kink.  This field opposes the current flowing in (slowing it down) and promotes the current flowing out (speeding it
  up) until these currents are equal, at which point there is no further
  accumulation of charge, and equilibrium is established.  It's a
  beautiful system, automatically self-correcting to keep the current
  uniform, and it does it all so quickly that, in practice, you can
  safely assume the current is the same all around the circuit, even in
  systems that oscillate at radio frequencies.

