Kim Y.

# How many grams of zinc metal are required to produce 10.0L of hydrogen gas at 25 degrees C and 1.00 atm pressure, by reaction with excess hydrochloric acid?

Zn (s) + 2HCl (aq) -> ZnCl2 (aq) + H2 (g)

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Brice T.

@J.R. S. Hello J.R.S. Thanks for your input. I understand your reasoning. However, the number of moles of zinc is equal to the number of moles of hydrogen gas, not half of it. According to the balanced reaction, the number of moles of zinc is the same as the number of moles of hydrogen gas, so moles Zn=0.409 moles, not 0.204 moles. The mistake that you made is calculating the number of moles using HCl, which is the excess reagent. You need to focus on Zn, the limiting reagent, because theoretically all of it is going to react to produce Hydrogen gas. Thus, in ideal case , all of the 26.7 g Zn are reacting to produce 10.0 L of Hydrogen gas. However, I know from experience in the lab that such a perfect reaction rarely occurs.
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J.R. S.

Of course you are absolutely correct. I don’t know what I was thinking. I guess I wasn’t thinking. This is basic stoichiometry and I simply goofed. Sorry.
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Brice T.

It is okay J.R.S. It happens to the best of us.
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J.R. S.

@Brice T. You forgot to use the correct stoichiometry where it takes TWO moles HCl for each ONE mole Zn. I believe the answer should be half of what you calculated it to be.
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