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At $10~^{\circ}\mathrm{C}$, a small piece of $\ce{Zn}$ reacts with $\mathrm{2.0~M}$ $\ce{HCl}$ to produce $\pu{15.0 mL}$ of $\ce{H2}$ gas in $\pu{25 s}$. What is the rate of the reaction?

My effort:

I know that the $\mathrm{2.0~M~} \ce{HCl}$ concentration needs to be used and started the following calculation:

$$\mathrm{2.0~M} \times \frac{1}{2} \times \frac{1}{\mathrm{25~s \times 0.015~ mL}} = \mathrm{0.00060~mol}~ \ce{H2}$$

$\mathrm{mol ~\ce{H2}} = \mathrm{mol~Zn} = \mathrm{0.0006~mol}$

However, the given answer is 0.060 mol of Zn.

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  • $\begingroup$ They are asking for a rate, so I am unsure about the given answer - I would expect it to be a function of time. Can you check the problem statement to be sure the information given is correct? $\endgroup$
    – Todd Minehardt
    Oct 31, 2015 at 19:51
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    $\begingroup$ one hint - So much much acid, so little hydrogen gas. $\endgroup$
    – MaxW
    Nov 1, 2015 at 0:51
  • $\begingroup$ The problem has two answers that can be solved depending on the interpretation of a "small piece of zinc". Assuming that there is excess zinc (like a zinc disc), and the overall reaction is limited not by contact area, then one answer is obtained. If you assume that the zinc is a sphere and (a) the contact area is decreasing and (b) the zinc is totally consumes at 25 seconds then you get another answer. If you assume that the contact area changes and that some zinc is left at 25 seconds then the problem can't be solved with the information provided. // I over think dumb problems like this. $\endgroup$
    – MaxW
    Nov 1, 2015 at 1:02
  • $\begingroup$ Assuming that there is excess zinc (like a zinc disc), and the overall reaction is limited not by contact area is the easiest way to solve the problem. Pick that one and state your assumptions in the solution. $\endgroup$
    – MaxW
    Nov 1, 2015 at 1:18
  • $\begingroup$ The question as stated is asking for a rate, but the given answer is a mass of Zn. Voting to close as unclear, due to this inconsistency. $\endgroup$
    – hBy2Py
    Jun 12, 2018 at 3:14

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