Plain S.

asked • 01/06/21

General Chemistry, Kind of Related to Stoichiometry, please help! really want to learn and understand this problem

This is a General Chemistry Question:

A reaction happens with 75 grams of acetic acid (C2O2H4) and an unknown metal, resulting in hydrogen gas and a solid metal with a mass of 113.4 g. What is the unknown metal?

Btw this was a question that I attempted but I'm stuck on a step where im getting that the unknown metal is 43.4 grams but Idk what the metal is since I need the answer to be in g/mol to find on the periodic table


J.R. S.

tutor
Just curious...how did you arrive at 43.4 g of metal?
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01/06/21

Robert S.

tutor
I second J.R. S.'s question. (113.4g - 75g) = 38.4g. You may have subtracted 70 grams by mistake? ( 113.4 - 70.0 = 43.4 grams).
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01/06/21

Plain S.

I calculated the number of grams of hydrogen based off of the number of mols of acetic acid. Since dividing 75 grams by the molar mass of acetic acid (60 g/mol) gives you the number of mols (1.25 g/mol) and we know that for every mol of acetic acid there are 2 mols of h2 so we multiply 1.25 by 2. So we have 2.5 mols of h2, and the molar mass of h2 is 2 g/mol. We multiply both these numbers and we get 5 grams of of h2. Now, we can plug everything back into the equation, and we get that the metal should be 43.4 grams. After converting this to amu (I used an online calculator) you get 26.1 amu, which is closest to aluminum.
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01/06/21

Plain S.

Also I accidentally wrote that the solid metal is 113.4 grams. It is not a solid metal, rather it is just a solid
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01/06/21

J.R. S.

tutor
Whoa...not solid metal, just a solid. How can you have a solid, unless it is unreacted metal. The products of the reaction are the metal acetate (aq) and H2(g). Where's the solid? Also, you get 1/2 mol H2 from each mol of acetic acid, not 2 moles H2. There's something wrong with this question, I'm afraid.
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01/07/21

Robert S.

tutor
I could assume the "solid" may be an precipitate of a metal salt. It's hard to answer the question unless we know the oxidation states of the reactants so that we can determine the molar ratios.
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01/09/21

1 Expert Answer

By:

J.R. S.

tutor
Dr. Bob... I did the same calculations but the problem is, as far as I know, there are no insoluble acetate salts. That’s why I asked “what solid”? My assumption was the metal was in excess, but I still don’t think there’s sufficient information to answer this, unless we’re missing something obvious.
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01/09/21

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