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Massive Karma for any Chem majors willing to help me

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607.
Posts: 521 - Karma: 35
I cant figure out this question on my General Chemistry Homework
Please Help

1. To a 150 mL beaker were added 0.510 g of solid copper powder and 2.5 mL of a 15 M aqueous solution of nitric acid. Brown fumes were evolved, and 25 mL of water was then added to the blue solution formed. To the aqueous solution was added 3.010 g of solid sodium carbonate in small portions. Only after all of the excess hydrogen ion in the aqueous solution was consumed by reaction with added carbonate ion did carbonate ion precipitate copper (II) ion as copper (II) carbonate. The solid copper (II) carbonate was isolated by vacuum filtration and dried; 0.920 g of copper (II) carbonate was collected. The net ionic equations for the reactions which occurred are shown below.

Cu(s) + 4H+(aq) + 2NO3-(aq) → Cu2+(aq) + 2NO2(g) + 2H2O(l)
2H+(aq) + CO32-(aq) → H2O(l) + CO2(g)
Cu2+(aq) + CO32-(aq) → CuCO3(s)

a. Calculate the percent yield of copper (II) carbonate.
b. Could copper turnings (i.e. thin, short ribbon of solid copper) be used instead of copper powder? Justify your choice.
c. Would using 3.010 g of solid sodium carbonate decahydrate instead of 3.010 g of sodium carbonate, with all other amounts of reagents being used kept the same as above, have any effect? Justify your decision, being as specific as possible.
Jan 19 2012 8:01PM
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JamesR
Posts: 3396 - Karma: -12
it should be criminal to teach science like that. chemistry is fascinating until it gets all of the life sucked out of it
Jan 19 2012 8:03PM
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Conker
Posts: 1379 - Karma: 6
for part a
for part b and c i think they just want you to talk about the speed of the reactions and surface areas?
Jan 19 2012 8:54PM
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Obi-Wan-Jenobi
Posts: 951 - Karma: 119
Ive done this exact lab. just take it slow step by step and youll be fine.
Jan 19 2012 8:57PM
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steep
Posts: 6063 - Karma: 297
I had my honors chem exam today. SO. HARD.
Jan 19 2012 9:09PM
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navierSTOKED
Posts: 858 - Karma: 33
This.  Part b is all about surface area.  The powder has a massively greater surface area thus the reaction will proceed ore rapidly.  If you use the turnings you risk the reaction taking so long that it is not at equilibrium when you take the readings.  Part c is about the concentration of the reactants.  If you use the carbonate in crystal form (ie. natron: decahydrate) the amount/mol of carbonate is actually less in your given mass thus you open up the same issues as b ie. the reaction proceeding too slowly.
Jan 19 2012 9:12PM
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607.
Posts: 521 - Karma: 35
Thank you +k to all
Jan 19 2012 9:21PM
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DrZoidberg
Posts: 5675 - Karma: 4,268
This one of those problems where you have to work backwards, right? Haven't done one of these in a while.

You know how much copper two carbonate you made, so you should be able to find out how much copper ion and CO3 were there to react to make that amount. You might aso be able to figure out how much CO3 reacted with the water from the second equation. Honestly I can't figure this out right now... Just closely inspect the order of reactions and the whole process and you'll see it.

But the powder versus solid, powder has more surface area and will react faster than a solid. So if the reaction runs it will be slower.

If the sodium decahydrate were relieved of it's hydrate component, it would interfere with the second reaction by adding products to the equation, shifting the equilibrium to the left. So it would be less favorable for that reaction to proceed. I haven't really learned about hydrate chemistry so that may be wrong. Maybe the H will react with the hydrate instead of the CO3?

Also vacuum filtration is annoying with powders if you're trying to get a high percent yield of a small amount of stuff.

Jan 19 2012 9:22PM
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Smokey.
Posts: 236 - Karma: 26
A- Change all your measurements to moles (good practice for upper level chem classes like orgo), then find the limiting reagent for the reaction, and using molar ratios, find your EXPECTED yield. Then find the moles of your EXPERIMENTAL yield. EXPERIMENTAL/EXPECTED x100= Percent Yield

B- The powder works the best, more surface area for more effective collisions.

C- Chances are, you were using Na2CO3 10H2O (Sodium Carbonate decahydrate) already and you just knew it as sodium carbonate. The decahydrate means is crystalized. Na2CO3 also come is 7H2O and 1H2O crystalline forms


Best of luck from a 2nd year Biochemistry major.

-Cody
Jan 19 2012 9:37PM
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unchyll
Posts: 4554 - Karma: 96
Jan 19 2012 9:38PM
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