Need help studying for Chem Test tommorow

[kabamawekesa]

Hi,

Im in Grade 11 Chemistry and I have a question. That I do not understand.

 

500g of copper metal is reacted with 2.5L of 3.0 mol/L nitric acid solution. Calculate how much of the copper metal remains after the reaction is complete.

 

I tried some ways but I still do not get the answer, I am looking for.

 

The answer is 320g of copper left unreacted.

 

Thank you,

KB

[hypervalent_iodine]

Well, what were the ways you tried?

 

The first step would be for you to write the reaction down and look at the stoicheometry. Were you able to figure that out?

[kabamawekesa]

Well, I was not sure which one to use since certain people online said to use one certain reaction .

 

The two reactions I got was,

 

Cu+4HNO3->Cu(NO3)2+2NO2+2H2O

or

HNO3+Cu->Cu(NO3)2+H2

 

which one is the correct one to use?

Hi,

I tried a method using Cu+4HNO3->Cu(NO3)2+2NO2+2H2O

The method I tried was, I used the concentration and volume of HNO3 to find the number of moles. ( 3.0x2.5)=7.5mol

then used the moles I found to find the moles of Cu(NO3)2. 7.5x 1moleCu(NO3)2/4HNO3= 1.875 moles

then took the moles I found with the Molar mass of CuNO3 to get the mass. (1.875x187.57= 351.7 grams).

I think I did something wrong,perhaps in finding the moles for Cu.

Edited December 14, 2012 by kabamawekesa

[hypervalent_iodine]

The first one is correct (the second one isn't even balanced).

You definitely need to go back and look at your concentrations. You've worked out the number of moles of nitric acid out fine, but you didn't work out the number of moles of copper metal. The copper metals is shown as Cu in the reaction, so you need to figure out how many moles there are in 500 g of Cu.

Once you have that, you need to look at which of your starting materials is the limiting reagent. As a hint, in your case you need to recognise that for every 1 mole of Cu that reacts, you need 4 moles of HNO₃. For instance, if you had 2 moles of Cu and 4 moles of HNO₃, only 1 mole of the Cu would react.

[kabamawekesa]

So,

since HNO3= 7.5 moles

and Cu=7.9 moles(500g/63.55), that means HNO3 is the limiting reactant right? Determined by the number of moles and since HNO3 has less moles, it is the limiting reagent.

But isnt there another way to find the moles of Cu? Can I take the moles found in HNO3 and multiply it by 1molCu/4molHNO3?

Giving me, 7.5mol*1molCu/4mol HNO3= 1.875 mol?

Edited December 14, 2012 by kabamawekesa

[hypervalent_iodine]

So,

since HNO3= 7.5 moles

and Cu=7.9 moles(500g/63.55), that means HNO3 is the limiting reactant right? Determined by the number of moles and since HNO3 has less moles, it is the limiting reagent.

 

That's right.

 

But isnt there another way to find the moles of Cu? Can I take the moles found in HNO3 and multiply it by 1molCu/4molHNO3?

Giving me, 7.5mol*1molCu/4mol HNO3= 1.875 mol?

 

Well sure, but you're finding out something different here. In this calculation, you're finding out the number of moles of Cu consumed in the reaction, whereas in the first one you were finding out the total number of moles of Cu are in the 500 g.

 

So, with this information, can you think of what to do next?

[kabamawekesa]

Uhm, No sorry. I have no idea what to do after that

[hypervalent_iodine]

The question is asking you how much copper remains after the reaction is complete. So if you know how many moles of Cu you have (~7.9 moles) and you know how many of those moles actually reacted (~1.89 moles), then you should be able to answer the question.

[kabamawekesa]

is it, 7.9-1.89=6.01 moles unreacted.

So, 6.01x 63.55g/mol= 382g left?

[hypervalent_iodine]

Looks right to me. I'm not sure why it differs from the answer you mentioned it was supposed to be, however. My only suggestion is that it might be a typo on their part (or I'm an idiot and I missed something), so perhaps check with your teacher.

[weiming1998]

Dilute nitric acid (3.0 mol/L would be dilute) reacts with copper in a different equation as the NO2 formed would redissolve back/oxidize the copper in solution:
3Cu(s)+8HNO3(aq)----->3Cu(NO3)2(aq)+2NO(g)+4H2O(l)

2.5L of the solution contain 7.5 moles of nitric acid (3x2.5). Multiply that by 3/8 (every 8 moles of HNO3 reacts with 3 moles of Cu) and you get 2.8125. 2.8125 moles of copper had reacted. Multiply that by 64 (atomic weight of copper) and you get 180. 180 grams of copper had reacted. 500-180=320.