Determination of % Comp of Al/Zn Alloy

  • #1

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
 

Answers and Replies

  • #2
GCT
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Write the relevant reaction equations displaying the products and reactants.
 
  • #3
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.
 
  • #4
Borek
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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?
 
  • #5
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?
 
  • #6
Borek
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No idea what 4.54 stands for. And note, that you are interested in TOTAL amount of hydrogen, from both reactions.
 
  • #7
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?
 
  • #8
Borek
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((.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.
 
  • #9
Um which details are faulty?
 
  • #10
Borek
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(x)(stoichiometric ratio)

But you did it OK in the other half of the equation, so thats probably just overlooked.
 
  • #11
Oh! Whoops my bad!
Thanks so much for your help, by the way!!
Much Appreciated :D
 

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