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$4.14$ 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.

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    $\begingroup$ You cannot produce more than maximum $m_\text{prod, max} = m_\text{reac} \cdot \frac{M_\text{prod}}{M_\text{reac}}$ , unless the product is not pure. But you can always produce less, if not all the lead formed the product or if there were loses. $\endgroup$
    – Poutnik
    Jul 6, 2023 at 9:54

2 Answers 2

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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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    $\begingroup$ Good explanation. Just to point out that you have taken mass as $4.41 \text{g}$ instead of $4.14 \text{g}$. $\endgroup$
    – Aleph
    Jul 6, 2023 at 11:29
  • $\begingroup$ @Aleph Thank you pointing out. The error in the equation (and subsequent result) was corrected. $\endgroup$
    – Buttonwood
    Jul 9, 2023 at 14:54
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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, 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. 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. "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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