Determination of % Comp of Al/Zn Alloy

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monkeysmine
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Homework Statement


The purpose of the lab is to determine the % comp of an alloy from the volume of hydrogen that is produced in the reaction.

We did a lab, in which an alloy was reacted with hydrochloric acid, in a test tube with a stopper and tube leading to a 500ml sidearm flask, with a clamp on the tubing. The sidearm flask was filled with water, and stoppered with a tube going into a 400ml beaker.

Assuming that the experiment was done correctly, the water from the flask was pushed into the beaker by the exothermic reaction.

DAta:
Pressure = 762 mmhg
Temp = 21.4C
Mass Alloy Trial One = .1229g, Trial Two = .1251g
Volume water produced T1 = 131.5ml, T2 = 140 ml.

Homework Equations


The Attempt at a Solution


I don't know where to start. =/

I calculated the amount of hydrogen produced, by calculating percent H in H20, and I got
14.61ml H2 gas for Trial 1, and 15.56ml H2 gas for Trial 2

I don't know where to go from here though.. any help would be appreciated
 
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Write the relevant reaction equations displaying the products and reactants.
 
Zn + HCl -> ZnCl2 + H2
Al + HCl -> AlCl3 + H2
if all I can think of..?
But I don't know how to calculate the amount of hydrogen produced. I know the hydrochloric must be the limiting reagent, to ensure complete reaction, but further than that, I'm stumped.
 
To be sure that reaction proceeded to the end, hydrochloric acid has to be in excess, sot it can't be a limiting reagent.

If I understand your description correctly, water was not pushed by 'exothermic' reaction, but by evolving gas. Volume of water displaced equals volume of produced hydrogen.

PV=nRT should give you number of moles of hydrogen.

Let's take a look at the first trial. Assume there were xg of Al in your sample. If so, there were (.1229g-xg) of Zn. Can you devise a formula that will let you calculate amount of hydrogen produced as a function of x?
 
Ummm...

Would (.1229-x gZn)(1mol/65.4gZn)(1/1) = 4.54
(x gAl)(1mol/27.0gAl)(3/2) = 4.54

be the right equations?
 
No idea what 4.54 stands for. And note, that you are interested in TOTAL amount of hydrogen, from both reactions.
 
Ah.. the five-day weekend has completely crippled my brain :(

I think I meant .00545, which is the moles of hydrogen produced from PV=nRT (762mmhg)(.1315L)/(62.4mmhg*L/moles*K)(294.4K).

I obtained .1315L as the amount of H2 produced, since volume water displaced = volume h2 produced.

Er. So if I let x = gAl, and .1229-x = gZn, then ((.1229-x)/MZn)(stoichiometric ratio) + (x)(stoichiometric ratio) = .00545 moles H2 produced, and solve for x?
 
monkeysmine said:
((.1229-x)/MZn)(stoichiometric ratio) + (x)(stoichiometric ratio) = .00545 moles H2 produced, and solve for x?

Apart from faulty details and unchecked numbers that's the correct idea.
 
monkeysmine said:
(x)(stoichiometric ratio)

But you did it OK in the other half of the equation, so that's probably just overlooked.
 
Oh! Whoops my bad!
Thanks so much for your help, by the way!
Much Appreciated :D