LarryLute
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Algebra to solve quantity; mixture of metals dissolved into HCl and H2 measured
There is a problem I can't solve. Please help me.
Here is the actual problem:
"2.00g mixture of aluminum and zinc metal powder is dissolved in HCl and 0.160g of H2-gas is produced. Calculate how much zinc and how much aluminum
there was. (mass)"
Stoiciometry is as follows:
2 Al + 6 HCl -> 2 AlCl3 +3 H2
Zn + 2 HCl -> ZnCl2 + H2
The solution manual produces the following equation:
Let the mass of Al be x; then mass of zinc is 2-x
1.5x/27 + 2-x/65.4 = 0.08
and says x is 1.2103
And that makes sence, but I end up getting x=2.03... 
because I get:
98.1x + 3531.6 -1765.8x = 141.264
And two websites I'v tried for help give negative numbers:
http://www.quickmath.com/webMathematica3/quickmath/equations...
http://www.wolframalpha.com/widget/widgetPopup.jsp?p=v&i...
Thanks! 
[Edited on 8-6-2014 by LarryLute]
[Edited on 8-6-2014 by LarryLute]
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Turner
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2=x+y
Al=x
Zn=y
y=2-x
therefore Zn=2-x
1Zn---> 1H2
2Al----> 3H2
(mass of Al)/(molar mass of Al) + (mass of zinc)/(molar mass of zinc) = 0.08moles of H2
since 2moles of Al create 3moles of H2 we need to multiply the mass of the Aluminum by 1.5 on left side, but this is really bothering me, why can't I
wrap my head around this?
2Al--->3H2
so shouldn't it be
2/3Al ---> 1H2?????
[Edited on 8-6-2014 by Turner]
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LarryLute
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Quote: Originally posted by Turner  | 2=x+y
Al=x
Zn=y
y=2-x
therefore Zn=2-x
1Zn---> 1H2
2Al----> 3H2
(mass of Al)/(molar mass of Al) + (mass of zinc)/(molar mass of zinc) = 0.08moles of H2
since 2moles of Al create 3moles of H2 we need to multiply the mass of the Aluminum by 1.5 on left side
(1.5x)/(27) + (2-x)/65.4 = .08 moles of H2
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Yes, but why do I get:
98.1x + 3531.6 -1765.8x = 141.264
x=2.03
?
I start to solve the equation by multiplying with the denominators.
[Edited on 8-6-2014 by LarryLute]
[Edited on 8-6-2014 by LarryLute]
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Turner
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just do (1.5 times 1/27)x + (2/65.4) - (x/65.4)
(2-x)/65.4 is the same as (2/65.4) - (x/65.4)
that may be where you went wrong
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aga
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Seems fairly simple, however i get a 'wrong' result every which way.
Rounding to sig figs helps, but still a different result.
Algebraically it appears to boil down to :-
2x = 3z
y = z
where x = moles Al, y = moles Zn, z = moles H2
0.160g H2 = 0.079 moles (3 dec places, as in the question).
so 4z (four being produced) = 0.079
z = 0.079 / 4 = 0.01975 = 0.020 to sig figs
x = ( 3 x 0.020 ) / 2 = 0.030
y = 0.020
Multiply by the aw to get the weights.
Calculating with 6 dec places ends up with 2.100228g total.
Rounding the mole results to 3 dec places, then multiplying = 2.117g total.
Rounding everything at every step gives 2.12g
Looks like one of either :-
a) the question is misleading (should say approx 2g mixed metals or not mention 2.00g at all).
b) i missed something vitally important in the Question
[Edited on 8-6-2014 by aga]
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blogfast25
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| Quote: Originally posted by aga |
Looks like the question is misleading (should say approx 2g mixed metals or not mention 2.00g at all).
[Edited on 8-6-2014 by aga] |
LOLROF. Why aga? Why is the person who poses the problem not allowed to be precise? That's BS.  (Scientist: 'Here, I've weighed 2.00 g of mixed metal'. Student: 'You can't do that: you can only weight it
approximately!'. Scientist: 'Why not?'. Student: 'Because otherwise the magix don't work!') 
Here:
Assume Al = x gram, so Zn = (2 - x) gram
x g Al is (x/27) mol Al. 2 mol Al would produce 6 g of H2, so (x/27) will produce (6/2) times (x/27) = x/9 g H2
(2 - x) g Zn is (2 - x)/65.4 mol Zn. 1 mol Zn would produce 2 g H2, so (2 - x)/65.4 g will produce 2(2 - x)/65.4 g H2.
Total H2 production was 0.160 g, so:
x/9 + 2(2 - x)/65.4 = 0.160
0.111 x + 4/65.4 - 2x/65.4 = 0.160
0.111 x + 0.0612 - 0.0306 x = 0.160
0.0804 x = 0.0988
Ergo x = 0.0988 / 0.0804 = 1.23 g Al
(2 - x) = 0.77 g Zn
Hydrogen production:
1.23/27 times 3 = 0.1366 g
0.77/65.4 times 2 = 0.0235 g
Total = 0.160 g H2
Top tip: to solve such problems set up the RIGHT algebraical equation!
Three variables with only two equations is by definition unresolved. That's why you NEED the 2.00 g total weight information.
Your third equation should have been: 27 x + 65.4 y = 2
[Edited on 8-6-2014 by blogfast25]
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aga
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Thanks for the corrections blogfast25.
Density of aga ? Very !
I went at it from a moles perspective rather than a weight persective, and the 2.00g in the Question was screaming at me not to do that, and i just
ignored it.
Classic 'did not read & understand the question properly' error.
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blogfast25
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It would have been perfectly possible to solve this in moles. But you just cannot ignore the total amount of metal because it relates to the total
amount of hydrogen released.
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aga
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So, took me a while, but here is solving for moles.
x = moles of Al, y = moles Zn
27x + 65.4y = 2.00
3x + 2y = 0.160
solving x in terms of y ...
(3x/2) * 2 + 2y = 0.16
= 3x + 2y = 0.16
= x = (0.16 - 2y)/3
substituting x for y term ...
27(0.16 - 2y)/3 + 65.4y = 2
= 9(0.16 - 2y) + 65.4y = 2
= 1.44 - 18y + 65.4y = 2
= 47.4y = 0.56
y = 0.011814 moles Zn
solving for x ...
27x + 65.4 * 0.011814 = 2
27x = 1.2274
x = 0.0455 moles Al
multiply by the a.w.s
Al = 0.0455 * 27 = 1.2285 grammes
Zn = 0.0118 * 65.4 = 0.7717 grammes
Yes, i like blogfast25's way much better as well.
(i know there are too many decimal places. It's Sunday, and beertime has begun, so reduce the sig figs in each step yourself !)
[Edited on 8-6-2014 by aga]
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blogfast25
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It's always a bit easier when you can formulate the problem algebraically in a single, one variable equation. But what you did works fine.
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aga
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Just thought it appropriate to give a different way to do it.
Took me ages, about 2 hours.
Had to check every single step.
Shows what age does.
Been at least 30 years since i last used that knowledge.
edit:
the thing that first bothered me was the relative reaction rates of Al and Zn qith HCl, and no time frame was cited.
By simply measuring the H2 weight, they all cancel out, and the result remains the same.
Neat.
[Edited on 8-6-2014 by aga]
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blogfast25
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| Quote: Originally posted by aga | the thing that first bothered me was the relative reaction rates of Al and Zn qith HCl, and no time frame was cited.
[Edited on 8-6-2014 by aga] |
It's implicit from context that the reactions have been allowed to run to completion (complete dissolution of all metal). The original problem would
have stated that, I think.
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aga
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| Quote: | | The original problem would have stated that, I think |
very likely. i was off in the realms of Uzn and Ual for a brief time due to that ommission.
Funny. Larry didn't even say thanks.
Was probly the chem teacher trying to Set the question then 
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