Chemistry: dissolving aluminum metal in hydrochloric acid

  • #1

Homework Statement


Aluminum metal dissolves in hydrochloric acid. What volume, in mL, of 1.58 M HCl is needed to completely dissolve 3.200 g of aluminum?(Hint: write the single-replacement reaction first)

Homework Equations


V[1]M[1]=V[2]M[2]

The Attempt at a Solution


I got the single-replacement reaction:
2Al(s) + 6HCl(aq) = 2AlCl[3](aq) + 3H[2](g)
But from there, I'm not really sure what to do. I think stoichiometry's involved, but I'm not sure how to use it.
 

Answers and Replies

  • #2
BvU
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If you have
2Al(s) + 6HCl(aq) = 2AlCl[3](aq) + 3H[2](g)
you have the stoechiometry ! My guess is that it's time to convert mol(ecul)es to mass :rolleyes:
 
  • #3
Borek
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My guess is that it's time to convert mol(ecul)es to mass
I would rather convert mass to moles :wink:
 
  • #4
BvU
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Sure, and perhaps you can explain to the audience why this is a better idea (generally or in this sspecific case) ?
 
  • #5
symbolipoint
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Stepwise path to find the volume of HCl,

Find moles of Aluminum (need formula weight of Aluminum)
Find moles of HCl (use ratio according to reaction)
Find volume of HCl solution of 1.58 M (concentration computation)
 
  • #6
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Am I correct in remembering that you may need to add a near-catalytic amount of copper chloride to accelerate breakdown of surface passivation ?

Or was this just for 'Activated Zinc' ??
 
  • #7
symbolipoint
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Am I correct in remembering that you may need to add a near-catalytic amount of copper chloride to accelerate breakdown of surface passivation ?

Or was this just for 'Activated Zinc' ??
Not in range. O.P. wants entry level course help, like for a beginning course. I am not saying that some kind of catalyst is not needed, but just that the idea is not in the range of the person's study.
 

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