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Topic: Aluminium liberation H2  (Read 9882 times)

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Offline huskywolf

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Aluminium liberation H2
« on: November 15, 2009, 01:14:56 PM »
Why does Al liberate more H2 from HCl than calcium?

Offline huskywolf

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Re: Aluminium liberation H2
« Reply #1 on: November 15, 2009, 01:31:40 PM »
Kitchen foil was put in HCl and liberated more H2 gas than calcium-HCl.is kitchen foil Aluminium oxide?is that why more H2 is produced

Offline Borek

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Re: Aluminium liberation H2
« Reply #2 on: November 15, 2009, 01:37:38 PM »
You were already explained twice in the previous thread why amount of metals matters, yet you are still repeating the same mistake and ask 'which metal gives more product'. This is a nonsensical question.
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Offline huskywolf

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Re: Aluminium liberation H2
« Reply #3 on: November 15, 2009, 01:48:10 PM »
the amount of Al 5g was the same amount of Ca 5g,I wan to know why the hell Al liberates ALOT more Hydrogen gas !You didnt explain that did you?

Offline huskywolf

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Re: Aluminium liberation H2
« Reply #4 on: November 15, 2009, 01:50:09 PM »
It must be something that is in the alu foil(kitchen foil)?the protective oxide layer or is there trapped hydrogen in the alu foil???

Offline Borek

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Re: Aluminium liberation H2
« Reply #5 on: November 15, 2009, 02:16:59 PM »
You didnt explain that did you?

No, but the question now - 5g vs 5g - makes sense.

Do you know whow to write reaction equations for both cases?

It must be something that is in the alu foil(kitchen foil)?the protective oxide layer or is there trapped hydrogen in the alu foil???

No, answer lies in the molar masses of both metals and their valences.
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Offline huskywolf

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Re: Aluminium liberation H2
« Reply #6 on: November 15, 2009, 02:51:11 PM »
yes
Ca + 2HCL=CaCl2 + H2
Al +  2HCL=AlCl2 + H2?

Offline huskywolf

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Re: Aluminium liberation H2
« Reply #7 on: November 15, 2009, 02:53:44 PM »
or should it be 2Al + 6HCl = 2AlCl3 + 3H2?

Offline Borek

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Re: Aluminium liberation H2
« Reply #8 on: November 15, 2009, 02:59:26 PM »
or should it be 2Al + 6HCl = 2AlCl3 + 3H2?

Much better now.

Looking at these equations, can you tell how many moles of H2 are produced per mole of Al? Per mole of Ca?
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Offline huskywolf

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Re: Aluminium liberation H2
« Reply #9 on: November 15, 2009, 03:09:09 PM »
2Al + 6HCl = 2AlCl3 + 3H2
2 mol al            :      3 mol Hydrogen gas   
Thank you!
1 more question

what are the ways I could determine the amount of acid consumed in a displacment reaction like these?

Offline Borek

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Re: Aluminium liberation H2
« Reply #10 on: November 15, 2009, 03:37:07 PM »
2Al + 6HCl = 2AlCl3 + 3H2
2 mol al            :      3 mol Hydrogen gas   
Thank you!

You have not answered my question - and at the moment we are still talking about moles, not masses, but if knowing that 2 moles of Al produce 3 moles of H2 makes you happy - that's OK with me. Just note it doesn't address your original problem.

Quote
what are the ways I could determine the amount of acid consumed in a displacment reaction like these?

The same way you can calculate amount of hydrogen produced - calculating number of moles and using stoichiometry as defined by the balanced reaction equation.
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Offline huskywolf

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Re: Aluminium liberation H2
« Reply #11 on: November 15, 2009, 03:59:55 PM »

sorry,ok so that would mean that more H2 gas would be liberated because 0.66 mole of Al would produce 1 mol of Hydrogen? so more hydrogen gas is produced than Ca would because Ca-H mole ratio is 1:1,so that would be the reason?

Offline Borek

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Re: Aluminium liberation H2
« Reply #12 on: November 15, 2009, 04:33:22 PM »
That would be the reason for equimolar amounts. For identical masses you should take into account also molar masses - 5 g of which metal means more moles of metal?
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Offline huskywolf

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Re: Aluminium liberation H2
« Reply #13 on: November 15, 2009, 04:43:23 PM »
That would be the reason for equimolar amounts. For identical masses you should take into account also molar masses - 5 g of which metal means more moles of metal

As Calciums atomic weight is 40 and Al's atomic weight is 27
n=m/M so n(al)=5/27=0.185 gmol
n(Ca)=5/40=0.125gmol

5g of Al =more moles of metal sir?

Offline Borek

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Re: Aluminium liberation H2
« Reply #14 on: November 15, 2009, 04:53:16 PM »
Yes.

And don't sir me :)
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