What is the meaning of "maximum" weight produced after series ot reactions?

$4.14$ grams of a piece of lead was made to react with $\ce{HNO3}$ so as to produce lead nitrate. Lead nitrate further reacts with a series of reagents to form $\ce{(NH4)2PbCl6}$. Then find the maximum possible weight in grams of $\ce{(NH4)2PbCl6}$ produced.


What is the meaning of maximum here? My method is to apply Principle of Atom Conservation on lead, but I directly get an answer without using the "maximum" condition.




The question can be answered assuming all atoms of lead ($\pu{4.14 g}$) end up in $\ce{(NH4)2PbCl6}$. So you need the molar masses of the two materials, because the number of moles of lead is the maximum number of moles of the product :


$$\frac{m(\ce{Pb})} {M(\ce{Pb})} = %
%
 \frac{m(\ce{(NH4)2PbCl6})} {M(\ce{(NH4)2PbCl6})}$$


$$\frac{\pu{4.14 g}} {\pu{207.20 g mol^{-1}} } = %
%
 \frac{m(\ce{(NH4)2PbCl6})} {\pu{455.99 g mol^{-1}}}$$



> 
>  Hence at maximum, you can isolate $m(\ce{(NH4)2PbCl6}) = \pu{9.11 g}$.
> 
> 
> 


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The question makes sense, but it is just needs explaining the use of "maximum" in this context.


Without overthinking the question, it asks what is theoretically the *maximum* yield by mass **assuming there is no loss of reactants or product**. Inevitably:


* Some of the raw material may be left (since there is an equilibrium, though it may be very far towards the side of the product).
* Some of the reactants and product may be rinsed away in the solvent or subsequent washings.
* Some of the product may be left on filter paper.
* For volatiles, such as some osmium compounds, some may be lost by evaporation.


The question "tries to make this easier for you," by positing 100% yield.


One could quibble with the question (though it likely won't get more points on a quiz).


* The wording, "maximum possible **weight**," rather than **mass**, means the product should be measured at the North or South Pole. Weight of a fixed mass [varies with position on Earth](https://en.wikipedia.org/wiki/Gravity_of_Earth), due to the Earth's spin, oblateness and altitude of the geographic location. There is a quite noticeable difference, ~0.7%, between weight of a mass at the equator and pole, which must be taken into account with spring and strain-gauge scales. At this writing, a 100 gram gold bar is priced at [US\$6,167.57](https://www.goldgrambars.com/). If it were sold by *weight* at the North Pole, it would have lost ~US$43 in value at the equator... but it ain't, it's sold by *mass*. Of course, a *balance* scale would read the same anywhere on Earth.
* On a *very* theoretical level, the (averaged) [atomic weight of lead varies by origin](https://en.wikipedia.org/wiki/Lead#Isotopes). "As time passes, the ratio of lead-206 and lead-207 to lead-204 increases," therefore "the standard atomic weight of lead is given to only one decimal place." So to calculate the *greatest maximum* (pardon, please, the pleonasm) yield of product, one could select lead from a location with the most 204Pb, the lightest stable isotope, giving the greatest number of moles of lead so that more of the other reactants are incorporated in the product.


*Raise these issues with your instructor at your own risk! Written by one who has been thrown out of class (in a kindly fashion) for arguing with a pedagogue.*


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