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Topic: Weight % of Zn / Al Mix  (Read 7980 times)

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Weight % of Zn / Al Mix
« on: April 22, 2004, 02:33:51 PM »

A 5.00-g mixture of zinc and aluminum reacted with HCL to give 1.278 x 10^2 ounces of hydrogen at 7.16 x 10^1 °F and 1.013 bar pressure. Th unbalance reactions are:

(a)  Zn + HCl ---> ZnCl2 + H2
(b)  Al + HCl ---> AlCl3 + H2

When balanced, the coefficients of reactants and products are:
(a) Zn + 2HCl --> ZnCl2 + H2
(b) 2Al + 6 HCl ---> 2AlCl3 + 3H2

Ok, this is where I have no CLUE....

(c) Calculate the percent by mass of zinc in the original mixture.
Now the hint that was given to me....I still have not clue ???
Hint -- % by mass zinc = mass of zinc/mass of mixture.

huh? ???  Maybe I am all chem out ...right now.... HELP?

OK.. UPDATE 4/23/2004. I think I have the answer to (c)
%Zn = 65.37/0.076 x 100 = 0.116 Zn
%Al = 26.98/0.041 x 100 = 0.151 Al

Can anyone check this for me and let me know if I am correct?
« Last Edit: April 24, 2004, 06:39:21 PM by hmx9123 »

Offline hmx9123

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Re:Zinc and Aluminum reacted with HCL
« Reply #1 on: April 22, 2004, 06:50:19 PM »
I think the original mixture refers to the mixture of Zn and Al, not of the HCl + Zn.  For that, you'd need to find out the mass of the zinc, and divide it by the total mass of the mixture of Zn and Al.  In your case, they already gave the mass of the mixture: 5.00g.  This means that:

% Zn = (Mass Zn) / (5.00g of Zn and Al)

(Times 100 of course, for the %)

So, first figure out what your mass of Zn is.  Can you figure out how many grams of Zn you have, given the equations and the other data you have?  If you need more help, send another message.

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