Why do we not using concentrated HCl and zinc reaction to prepare hydrogen? First apologize for this kind silly question.One of my student ask me this question but I was unable to get a perfect answer from internet.Generally dilute HCl and impure zinc are using for hydrogen production in laboratory.My question is why  do reaction stoped after some little time reaction if we use pure zinc and concentrated HCl?
Is it for coverup zinc by zinc sulphate?
It will be thankful for any kind help.
If metallic zinc $\ce{Zn}$ is in contact with concentrated $\ce{HCl}$ solution, $\ce{H2}$ is produced, but the reaction is exothermic  : the solution gets hot. As $\ce{HCl}$ is not so highly soluble in hot water, a fraction of the dissolved  $\ce{HCl}$ will be vaporized. So the gas produced will be a mixture of $\ce{H2}$ and $\ce{HCl}$, which is not wanted.
Anyhow, the reaction between pure $\ce{Zn}$ and pure concentrated $\ce{HCl}$ is not really rapid. And, after some time, when the solution contains a certain amount of $\ce{Zn^{2+}}$, the reaction rate decreases in such a manner that the reaction looks finished. It is rather surprising, and even deceiving.
However, if some metallic impurities are present in solution, like $\ce{Co^{2+}}$ or $\ce{Cu^{2+}}$, the zinc metal reduces these ions according to $$\ce{Zn + Co^{2+} -&gt; Co  +  Zn^{2+}}$$or $$\ce{Zn + Cu^{2+} -&gt; Cu  +  Zn^{2+}}$$ As a result, some small amounts of metallic cobalt or copper are deposited on the zinc metal. The contact of the two metals $\ce{Zn + Co}$ or $\ce{Cu}$ produces a galvanic cell, and the reaction rate increases a lot. $\ce{Zn}$ gets dissolved much quicker, producing electrons that go onto the copper or cobalt spots, where they reduce $\ce{H+}$ to produce a lot of $\ce{H2}$.
So the best way of producing H2 by the reaction $\ce{Zn + HCl}$ is to use not too concentrated solutions of $\ce{HCl}$ solution, with small amounts of copper or cobalt salts added into the acidic solution.
The same phenomena of galvanic cell happens if  $\ce{HCl}$ is replaced by $\ce{H2SO4}$