$\pu{9.547 g}$ $\ce{Cu}$ tile is added to an $\ce{AgNO3}$ solution. After some time, $\ce{Cu}$ was taken out of the solution, washed, dried and weighed. The mass appeared to be $\pu{9.983 g}$. How much of $\ce{Ag}$ has formed on the $\ce{Cu}$ tile?

The right answer to be found is supposed to be: $\pu{1.08 g}$.

My imaginary equations (with unlikely scenarios):

\begin{align} \ce{Cu + 2AgNO3 &-> Cu(NO3)2 + 2Ag}\\ \ce{Cu + AgNO3 &-> CuNO3 + 2Ag} \end{align}

I cannot solve this problem. There are a few others, but this is bugging me the most. How does one solve it? I believe this is something I wasn't taught before or there is again a mistake in the references given. I've tried calculating in ways I thought was right and anyhow, I cannot find the answer.

After discussions, I have come to the conclusion that I am taking $\pu{0.618 g}$ as the solution to this problem and will look forward into it tomorrow with the teacher as well.
This is the only closest answer I seem to arrive at by myself and others, as we get $\pu{0.618 g}$ when using the $\Delta m_\mathrm{p}=\pu{0.436 g}$ and $\Delta m_\mathrm{t}=\pu{22.853 g}$ differences. However, the problem regarding the other exercises still persists, which are of similar nature. I am starting to question my teacher's logic and abilities.

There is a close answer to the referenced answer, which assumes the $\ce{Cu}$ having a charge of $+1$, thus $\ce{Cu + Ag+ -> Cu+ + Ag}$ and $\Delta m_\mathrm{p}=\pu{0.436 g}$, $\Delta m_\mathrm{t}=\pu{16.2 g}$. However, I assume that to be a very unlikely scenario considering the current circumstances. This scenario lets you acquire \pu{1.0617 g} as the answer.

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    $\begingroup$ Could you please edit what you've tried so far into the question? $\endgroup$ Commented Feb 23, 2019 at 13:39
  • $\begingroup$ To me it seems like there is a typo somewhere. For some reason this question is all over Runet (Russian internet) and the answer is 0.62 g (which to me is the correct one). $\endgroup$
    – andselisk
    Commented Feb 23, 2019 at 14:02
  • $\begingroup$ Could you provide a link for that? I can't seen to find anything regarding this anywhere $\endgroup$ Commented Feb 23, 2019 at 14:55
  • $\begingroup$ @Alchimista This is my first post on stack, I got extremely surprised by the hold put on the topic, as well as it being treated as off-topic. I even put the homework tag beforehand, yet it was removed by the moderator $\endgroup$ Commented Feb 24, 2019 at 19:40
  • $\begingroup$ I have reopened the question, but I'd still think it would have been a much better question, if the actual calculation that you did were included. In its current form it only discusses results, not the way to get them, therefore it doesn't point to the actual question you have. You should also explain the abbreviations you use. This question was probably closed, because there was a certain lack of attempt in previous versions. the edits then have not reached the critical number to reopen. (Please note that the homework tag is deprecated, it says that in all caps in its description.) $\endgroup$ Commented Mar 2, 2019 at 14:01

2 Answers 2


In this experiment the tile lose mass due to oxidation and dissolution as Cu dication (Cu++) and acquire mass due to the reduction and deposition of Ag.

This is what is measured by weighting before and after.

This difference in mass correspond to 2n moles of Ag minus n moles of Cu so is just matter of solving a first degree equation.

Quickly, without being consistent in digits and truncation., solving it for the current data gave about 0.6 g of Ag.

All others outcomes seems to me to assume an unlike stoichiometry or refuses perpetuating around.

A way to get into an "effective stoichiometry" of 1:1 is to consider formation of Cu+ followed by disproportionation AND assuming that all metal copper finely precipitates without depositing back to the tile from which originally came from. While, partially occurring, this scenario would explain discrepancies found performing a real experiment, it seems unlikely that is intended as to give a fixed result as for it requires the assumption in italic above. Doing that the amount of silver is now about 1.06 g.

  • $\begingroup$ Thank you for the answer. Can you possibly provide how you got the answer and why 1.08g is not a viable answer? I wish to discuss this with the teacher whether she got the wrong answer herself $\endgroup$ Commented Feb 23, 2019 at 21:09
  • $\begingroup$ You get it exactly as I wrote assuming that tge reaction is Cu and 2 Ag+ giving 2 Ag and Cu++. It seems the likely one. You can play with 1 to 1 stoichiometry but I guess you did and does not give the answer. If not retry following the same reasoning as in my answer. It would be funny if it turns to be the wanted 1.08 g :)) but what I and and others have proposed (Cu++) is the one that makes sense. $\endgroup$
    – Alchimista
    Commented Feb 23, 2019 at 21:29
  • $\begingroup$ It seems given by an author that thought of that. But it could be a mistake. You could add another exercise of that sort to see if it is the same problem. Read again my answer as I added an end. But my opinion is unchanged. $\endgroup$
    – Alchimista
    Commented Feb 24, 2019 at 0:17
  • $\begingroup$ Another exercise: 73g of Zn put in NiSO4 solution, which has a mass of 240g. After some time, the Zn tile appeared to be 71.8g. Find ZnSO4 mass percent in the solution (Answer: 13.35%) $\endgroup$ Commented Feb 24, 2019 at 0:19
  • $\begingroup$ Thank you. If you can, could you provide the steps if you manage to get the answer? (possibly in chat or somewhere) As well as for the previous exercise where you get 1.06g, as that seems to be the most accurate answer I've yet to see and never gotten, as I am completely frustrated with just these exercises $\endgroup$ Commented Feb 24, 2019 at 0:22

Well, let's work backwards... Given:

9.547g Cu tile

9.983g tile + Ag

1.08 g Ag

then the Cu that is left on the tile is

9.983 - 1.080 = 8.903 g

and the Cu that dissolved must be

9.547 - 8.903 = 0.644 g

Using the atomic masses:

AM(Cu) = 63.546

AM(Ag) = 107.8682

we can calculate the moles of Ag

1.08/107.8682 = 0.010012

and the moles of Cu

0.644/63.546 = 0.010134

So the poser of the question was assuming the chemical reaction:

$\ce{Cu + Ag^+ -> Cu^+ + Ag}$

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    $\begingroup$ Copper(I) in aqueous solution? No way. $\endgroup$
    – andselisk
    Commented Feb 23, 2019 at 15:08
  • $\begingroup$ Didn't say that I agreed, just what poser of the question must have done... $\endgroup$
    – MaxW
    Commented Feb 23, 2019 at 15:09
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    $\begingroup$ Charge balance is fubar for that one. $\endgroup$
    – andselisk
    Commented Feb 23, 2019 at 15:29
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    $\begingroup$ All in all I don't know how the poser messed the problem up. Old problems get reworked to form new problems. Perhaps in the original problem there was some sort of metal complexes that gave the overall 1:1 reaction. $\endgroup$
    – MaxW
    Commented Feb 23, 2019 at 15:29
  • 1
    $\begingroup$ @I'mPatrick - Lacking any other information I'd expect an acidic solution and that the copper would be oxidized to $\ce{Cu^{2+}}$. However to get the given answer you have to reason that the copper goes to $\ce{Cu^+}$ and assume some small rounding errors. $\endgroup$
    – MaxW
    Commented Feb 23, 2019 at 17:20

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