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### Topic: Need help with silver nitrate/copper lab...  (Read 11618 times)

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#### Noob

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##### Need help with silver nitrate/copper lab...
« on: September 03, 2005, 08:58:35 PM »
I'm doing a writeup on my silver nitrate and copper wire lab and I don't know if I'm doing the right steps or not, so any help would be greatly appreciated.

First of all the balanced equation is:
2AgNO3(aq) + 1Cu(s) ---> 2Ag(s) + 1Cu(NO3)2(aq)

So my theoretical yield is:
339.80g AgNO3 + 63.50g Cu = 215.70g Ag + 187.50g Cu(NO3)2

The data I got was:
9.860g AgNO3 + 3.760g Cu = 5.935g Ag + 1.945g Cu(NO3)2

So 9.86g AgNO3 x 1 mol AgNO3 / 169.9g AgNo3 = 0.058 mol AgNO3. Because I used 29ml of distilled water my molarity would be 0.058 / 0.029 = 2 M. I'm not sure if this is right but 0.058 mol AgNO3 x 107.86g Ag = 6.256g Ag should be my expected amount of silver after the reaction correct? From here I'm not sure what do do with molarity or what else to do with the amount of moles.

I also did the percent compostition: 9.86g AgNO3 x 107.76g Ag / 169.86g AgNO3 = 6.26g Ag and 3.76g Cu x 63.5g Cu / 187.5g Cu(NO3)2 = 1.273g Cu. I'm not sure if this is correct but I thought that 6.26g and 1.276g should be the expected amount of silver and copper after the reaction.

So my percent yield should be: 5.935g Ag / 6.26g Ag = 95% and 9.86G AgNO3 / 339.8g AgNO3 = 2.9% x 187.5 = Cu(NO3)2 = 5.44g Cu(NO3)2 so 1.945g Cu(NO3)2 / 5.44g Cu(NO3)2 = 36%. Is this correct?

I'm really confused on whether or not I'm doing the right steps or right calculations, mainly the amount of moles and molarity is what I don't know what to do with. Thanks for your help.

#### xiankai

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##### Re:Need help with silver nitrate/copper lab...
« Reply #1 on: September 03, 2005, 09:47:41 PM »
Quote
The data I got was:
9.860g AgNO3 + 3.760g Cu = 5.935g Ag + 1.945g Cu(NO3)2

based on the law of conservation of mass, i feel that there's something wrong. can u explain how u got the values, step by step?

Quote
Because I used 29ml of distilled water my molarity would be 0.058 / 0.029 = 2 M. I'm not sure if this is right

have u factored in the amount of liquid present in Cu(NO3)2?

Quote
I also did the percent compostition: 9.86g AgNO3 x 107.76g Ag / 169.86g AgNO3 = 6.26g Ag and 3.76g Cu x 63.5g Cu / 187.5g Cu(NO3)2 = 1.273g Cu. I'm not sure if this is correct but I thought that 6.26g and 1.276g should be the expected amount of silver and copper after the reaction.

the proccess seems right to me.

Quote
9.86G AgNO3 / 339.8g AgNO3 = 2.9% x 187.5 = Cu(NO3)2 = 5.44g Cu(NO3)2 so 1.945g Cu(NO3)2 / 5.44g Cu(NO3)2 = 36%. Is this correct?
it should be 0.029 moles, because thats what u're calculating. continuing onwards, i see nothing else wrong with your calculations.

but have u factored in that reageants may be in excess? that some unreacted reageants may be left over? im curious to see how u obtained the data u obtained.
one learns best by teaching

#### Noob

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##### Re:Need help with silver nitrate/copper lab...
« Reply #2 on: September 03, 2005, 10:22:34 PM »
I didn't actually do the lab because I got switched into the class after it had already started so I just used another person's results for my lab. I noticed the data seemed really off as well, but this is the data I have to use. The procedure is silver nitrate into the distilled water, copper wire coiled and into the silver nitrate, overnight, shake off the silver on wire, pour through filter paper, dry silver and weigh it. I think the reason why the data is so inaccurate is because the copper wire they used was too thin and broke under the weight of all the silver on it. If that were the case that would make copper the limited reagent, meaning alot of the silver ion didn't react. I'm not sure what happen though cause I wasn't there to see it.

#### xiankai

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##### Re:Need help with silver nitrate/copper lab...
« Reply #3 on: September 04, 2005, 12:29:19 PM »
in that case, the only reliable data we have is copper and silver : silver nitrate weighed has some water, copper nitrate too.

what does the question ask for?
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