Mole Ratios: Silver Nitrate + Copper (Equation)

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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

2. You almost have the balanced equation already, except you got the formula for copper(II) nitrate wrong.



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:




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



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.




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



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.

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)

Ok, I'll continue it if you allow me to do so.

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

So,



Then in whole numbers;



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!

You'll see that the coefficients of your equation are not the same as that you predicted, that is


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 !

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.

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?)

Copper solutions are a pale blue. The copper ammonia complex that I *thought* you had is an intense blue.