# Mole Ratios: Silver Nitrate + Copper (Equation)

[F][/font]
\

Silver Nitrate + Copper

AgNO3 + Cu ---> CuNO3 + Ag

The reaction starts out with

Weight of empty beaker - 111.420

200ml of Silver Nitrate with a weight of 316.262g(including beaker)

added 1g of Copper weight 317.262g(including beaker)

After given amount of time:

Decanted solution into empty beaker2 weight is now 313.868

In beaker 1 Silver remains weight is 114.815

A.) 1 gram of copper

B.) 114.815g silver and beaker

C.)empty beaker 111.420g

D.)3.395g of silver produced

(e) Moles of solid copper used in reaction (mol)
1g x 1mol/63.546 = .016 mol

(f) Moles of solid silver produced in reaction (mol)
3.395 x 1 mol/107.868 = .031 mol

I have gotten this far but this is the rest of my assignment and I am stuck when it comes to the mole ratios and fractional coefficients.

2. Write the equation for the reaction between copper and silver ion. Include your experimentally determined mole ratios as fractional coefficients.

3. Convert the fractional coefficients to a whole number ratio and rewrite the equation using the whole number ratio.

4. If the silver in the beaker contained water during your last weighing, how would this affect your results?

5. Assume that magnesium would act atom-for-atom exactly the same as copper in this experiment. How many grams of magnesium would, have been used in the reaction if one gram of silver were produced? The atomic mass of magnesium is 24.31 g/mol.

6. Explain the source of the blue color of the solution after the reaction.

 Curlyben Posts: 18,081, Reputation: 8728 Admin & Wine Expert #2 Apr 19, 2009, 09:50 PM
Thank you for taking the time to copy your homework to AMHD.
Please refer to this announcement: Ask Me Help Desk - Announcements in Forum : Homework Help
 Pre-KTiff Posts: 6, Reputation: 5 New Member #3 Apr 19, 2009, 10:17 PM
OK, I am starting to regret that I even joined this place because as I have been looking around for some help to help me better understand this all I have seen is people telling people to do their own work. DUH! That is why we are here! We need help. I need someone to help me to explain it to me so that I may better understand. NOW; if I wanted someone to give me the answer I know I could find a classmate that would tell me but instead I make an effort to try and seek help on the web where I WAS sure to find someone who could explain the meaning to me. NOW, the reason I wrote in my question is so my viewers could better understand what I am needing to accomplish.

SO, First starters, why don't YOU PEOPLE start by leading people in the right direction instead of telling them what they are constantly doing wrong. That is what makes people so afraid to even ask a question.

I just simply said that I am stuck, I didn't say CAN SOMEONE PLEASE DO THIS QUESTION OR ANSWER IT FOR ME? Instead I asked for help!

NOW! Now that you know, I am here for help, can someone please steer me in the right direction help me to understand how to determine the mole ratios.

Furthermore, I have done most of the work. All of the stuff at the beggining I have already done in my online lab simulation. It is just Chemistry, there is nothing about Chemistry that is short and simple.

Thanks Tiffany
 Perito Posts: 3,141, Reputation: 781 Ultra Member #4 Apr 20, 2009, 05:15 AM
Quote:
 Originally Posted by Pre-KTiff AgNO3 + Cu ---> CuNO3 + Ag Weight of empty beaker - 111.420 200ml of Silver Nitrate with a weight of 316.262g(including beaker) Added 1g of Copper weight 317.262g(including beaker) After given amount of time: Decanted solution into empty beaker2 weight is now 313.868 In beaker 1 Silver remains weight is 114.815 A.) 1 gram of copper B.) 114.815g silver and beaker C.)empty beaker 111.420g D.)3.395g of silver produced (e) Moles of solid copper used in reaction (mol) 1g x 1mol/63.546 = .016 mol (f) Moles of solid silver produced in reaction (mol) 3.395 x 1 mol/107.868 = .031 mol 2. Write the equation for the reaction between copper and silver ion. Include your experimentally determined mole ratios as fractional coefficients. 3. Convert the fractional coefficients to a whole number ratio and rewrite the equation using the whole number ratio. 4. If the silver in the beaker contained water during your last weighing, how would this affect your results? 5. Assume that magnesium would act atom-for-atom exactly the same as copper in this experiment. How many grams of magnesium would, have been used in the reaction if one gram of silver were produced? The atomic mass of magnesium is 24.31 g/mol. 6. Explain the source of the blue color of the solution after the reaction.
2. You almost have the balanced equation already, except you got the formula for copper(II) nitrate wrong.

$2AgNO3 + Cu ---> Cu(NO_3)_2 + 2Ag$

You could ignore the Nitrate since it's just a "spectator ion". I'm going to write this using "half reactions" to show you how that can be used to balance equations rather easily:

$Ag^+\, +\, e^\,--->\, Ag$
$Cu\,--->\, Cu^+2\,+\,2e^-$

Multiply the top reaction by 2 to balance the electrons and directly add the half reactions to get the whole reaction:

$2Ag^+\,Cu\,=\,Cu^+2\,+\,2Ag$

3. Unfortunately, I don't understand this. What fractional coefficients are they talking about? Maybe it's this equation being converted to the above equation.

$Ag^+\,\frac 12 Cu\,=\,\frac 12 Cu^+2\,+\,Ag$

4. If the silver were not dry, the weight would be too high. This would lead to an error.

5. You know the number of moles of copper. Multiply the number of moles of copper by the atomic weight (atomic mass) of magnesium (24.31). That's the number of grams of magnesium that would have been used.

6. The blue color? Copper forms blue complexes with ammonia

$Cu(H_2O)_6^{+2}$

You didn't mention anything about ammonia, but that's where it has to come from.

Sometimes people simply copy the problem to the page. CurlyBen likes to enforce the rule that you should tell us what you've done and we'll help. You did that. Sorry you didn't get an earlier response.
 Pre-KTiff Posts: 6, Reputation: 5 New Member #5 Apr 20, 2009, 09:55 AM
First, I would like to thank you so much for your time. It means a great deal to me.

Secondly, I like the way you have laid Q #2; it makes it more simple to understand; Why can't the instructors explain it that simple?

Ok, I still don't get this. My mole ratios are as follows:
Copper = .016; if this were turned into a fraction it would be 16/100

Silver = .031; if it were fraction 31/100

"Include your experimentally determined mole ratios as fractional coefficients."

Also, there is nothing in this lab assignment that requires ammonia but I did find that Copper turns into a blue aqueous solution when it reacts with the silver nitrate.

5.) 0.016 mol x 24.31 = 0.389g (did I do that right)
 Unknown008 Posts: 8,147, Reputation: 3745 Uber Member #6 Apr 20, 2009, 10:28 AM
Ok, I'll continue it if you allow me to do so.

You obtained the fractions as 16/100 and 31/100

So,

$\frac{31}{100}AgNO_3 + \frac{16}{100}Cu ---\rightarrow \frac{16}{100}Cu(NO_3)_2 + \frac{31}{100}Ag$

Then in whole numbers;

$31AgNO_3 + 16Cu ---\rightarrow 16Cu(NO_3)_2 + 31Ag$

If I'm not mistaken, copper (II) nitrate is slightly blue. Perhaps you won't notice it just by looking at it but if you place a blank white page behind a beaker containing copper(II) nitrate, you'll see the difference.

5. Yup you did it well. Do you know how you got there and why multiply?

Look here, you had 0.016 moles of copper reacting, k? Then, Mg act atom for atom as copper. Therefore, if one atom of copper reacted, one atom of Mg will react.

The same way round, if 0.016 mol of Cu react, 0.016 mol of Mg will react. Got it?

But Mg has a different mass compared to Cu.
0.016 mol of Cu has a mass of (0.016 x 65.4)g
0.016 mol of Mg has a mass of (0.016 x 24.3)g

Hope it helped!
 Unknown008 Posts: 8,147, Reputation: 3745 Uber Member #7 Apr 20, 2009, 10:31 AM
You'll see that the coefficients of your equation are not the same as that you predicted, that is
$2AgNO_3 + Cu ---\rightarrow Cu(NO_3)_2 +2 Ag$

But you'll see that they are quite close...

Mole ratio of AgNO3 : Cu = 2 : 1
Mole ratio you got = 31:16 which is approximately 2:1 !
 Pre-KTiff Posts: 6, Reputation: 5 New Member #8 Apr 20, 2009, 12:14 PM
Oh, Ok, I see. So actually if I hadn't of rounded the number off it would have been on the money,

Thanks so much.
 Pre-KTiff Posts: 6, Reputation: 5 New Member #9 Apr 20, 2009, 12:18 PM
You said why multiply, this is what I think,
You should multiply because by finding the mole you divide (ex. 1/24.31) then you multiply that by the amount of grams you have to find the amount of moles in the grams. (am I right?)
 Perito Posts: 3,141, Reputation: 781 Ultra Member #10 Apr 20, 2009, 12:26 PM
Quote:
 Originally Posted by Pre-KTiff Also, there is nothing in this lab assignment that requires ammonia but I did find that Copper turns into a blue aqueous solution when it reacts with the silver nitrate.
Copper solutions are a pale blue. The copper ammonia complex that I *thought* you had is an intense blue.

## Check out some similar questions!

?

Silver nitrate and skin tags [ 1 Answers ]

Hi, Can you use silver nitrate sticks to remove small skin tags?

Mole-mole ratios [ 6 Answers ]

:confused: Given the fallowing balanced equation, 2Al + 3S -> Al2S3 Q 1. How many moles of Al2S3 will form if we start with 16 moles of Al? Q 2. How many moles of sulfur will be combined with 16 moles of Al? ...