For calculating overall yield of a multistep (linear) reaction why do we multiply the yield of each step (and then multiply by 100)?

Why not just take average of the yields of each step and find the percent?

What's the logic?

@Abcd when we say the yield of the n-th step is A%, that means the moles of the product produced in n-th step is A% of that step's reactant. So, say you had X molecules of the first reactant. Assume the successive yields of a three-step reaction are 75%, 60%, 30%. Here, note that the reactant of n-th step would be the product of the (n-1)-th step. So, the yield of the final product would be = 30 % * product 2 = 30% * (60% * product 1) = 30% * 60% * (75% * reactant) = 0.3*0.6*0.75*X

Could anyone confirm if the answer here is $K_c = 1$?

Thanks @hBy2Py :D

There is/was/will be always some ambiguity when it comes to chemistry involving illicit substances (Re: If use HCOOC or HClO4 for redcution formetamide, will I have higer yeild amphetamine sulpate?) about how much they are on topic on our site. Should we have a more detailed discussion about it? (I closed the example mainly because of the multitude of questions involved as too broad.)

@Martin-マーチン That was VLQ altogether, could be even deleted immediately IMO

You lured me in, @Gaurang

@hBy2Py thanks so much :D I now feel I should have put a higher bounty on that, the amount of effort you put into it is absolutely superb!

the question was well-presented. But it didn't get much attention, which I believe was because it was way too long. I am glad you took the time to answer it :D

@Martin-マーチン It's things like chemistry.stackexchange.com/questions/35847/… that could lead to discussion.

@Mithoron yes indeed, by VLQ is no close reason, hence to broad.

@Mithoron I'm not sure; I don't see that particularly problematic. The information is already out there anyway. I simply don't understand the fascination with such subjects.