What Volume of Copper Sulfate Solution is Needed for 1.110g of Iron?

[kirsten_2009]

URGENT Chemistry Stoichiometry Help :S

Homework Statement

Suppose that you use 1.110g of iron in this experiment. What is the minimum volume of 1.0M copper sulfate solution that you should add?

Homework Equations

Fe(s) + Cu2+(aq) --> Fe2+(aq) + Cu(s) (1)

2Fe(s) + Cu2+(aq) --> 2Fe3+(aq) + 3Cu(s) (2)

*****Since I do not know at this stage whether equation (1) or (2) is the appropriate one, I must base my calculation on whichever of the equations would require the most copper sulfate for a given quantity of iron*****

The Attempt at a Solution

Here is what I believe to be a very poor attempt at solving the problem...any help will be greatly appreciated!

1.110g of Fe x (1 mol / 59.0 g Fe) = 0.018813559 moles Fe

0.018813559 mol Fe x (3 mol Cu2+ / 2 mol Fe) = 0.028220338 mol Cu2+

AND this is where I become stuck since the answer I need uses litres but all I have are the moles of Cu2+. Also, I don't know where the whole 1.0M from the question comes into play. I really need help. Thanks in advance!

 

[Borek]

Fe²⁺ it is.

$$C = \frac n V$$

Do you know how to solve for n?

 

[kirsten_2009]

Hello! First of all, thanks so much for replying to me! I am in dire need of help.

C = n/v makes sense. I would solve for "V", I have "C" (= 1.0 M) and "N" is the number of moles, correct?

Now, solving for "n" would give me the moles of copper sulfate? Do I obtain this number by using molar ratios? If so, how would I go about doing that? I think I solved for the moles of Cu2+ above but I don't know how I would get the moles of copper sulfate...HELP! :)

 

[Borek]

You have used wrong reaction equation so your result was off, but the approach was correct. Just note that molar mass of iron is not 59.

If there is one atom of copper per one molecule of copper sulfate, how many moles of copper sulfate per one mole of copper atoms (or copper ions)?

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methods

 

[kirsten_2009]

Again, thanks for helping me out. Also, I like how you don't just give me the answer, you "walk me through it" which is actually what I need, better learning experience!

So...

Using the correct reaction equation and the correct molar mass for iron now:

1.110 g Fe x (1 mol / 56.0 g Fe) = 0.018813559 mol Fe

0.018813559 mol Fe x (1 mol Cu2+ / 1 mol Fe) = 0.018813559 mol Cu2+

and the ratio between moles of Copper Sulfate and copper atoms is 1:1

-Since the question is asking the volume of copper sulfate needed, I'm assuming my answer should be in litres? Is that where the c=n/v formula comes in? would this be correct?

C = n/v ------> V= c x n

V = 1.0M x 0.018813559
V = 0.018813559 L

Somehow this doesn't seem right...I'm so sorry to bother you but I'm just not good at chemistry no matter how hard I try and I have nobody else to help me. Thanks a lot!

 

[Redbelly98]

Borek may be logged off for the night, as he lives in Europe, so I'll try to help out until he returns.

Using the correct reaction equation and the correct molar mass for iron now:

1.110 g Fe x (1 mol / 56.0 g Fe) = 0.018813559 mol Fe

It looks like you still used 59 when you did this calculation. Redo it using 56.

0.018813559 mol Fe x (1 mol Cu2+ / 1 mol Fe) = 0.018813559 mol Cu2+

and the ratio between moles of Copper Sulfate and copper atoms is 1:1

-Since the question is asking the volume of copper sulfate needed, I'm assuming my answer should be in litres? Is that where the c=n/v formula comes in? would this be correct?

Yes.

C = n/v ------> V= c x n

Actually, doing the algebra gives ---> V = n/C
But, since C=1.0M, the numbers work out the same!

V = 1.0M x 0.018813559
V = 0.018813559 L

Actually V = 0.01881 mol / 1.0M = 0.01881 L
or round it off to 0.019 L, since the 1.0M is given to just 2 significant figures.

Somehow this doesn't seem right...I'm so sorry to bother you but I'm just not good at chemistry no matter how hard I try and I have nobody else to help me. Thanks a lot!

Looks pretty close to me , but you'll need to redo it using the correct 56 g/mol for iron.

 

[kirsten_2009]

Hello Redbelly! Oh my! Yes, you are right! I did use the 59 again!

THANK YOU SOOOOOOO MUCH! I really appreciate all the help from both Borek and yourself! You guys have made all the difference for me!

I never took high school chemistry, but now I need to pass first year university chemistry so that I can graduate this summer. Basically, I just taught myself (not very well by the way) in the last 3 months all of high school chemistry and now I'm taking this university course and it's so difficult for me! Anyways, thanks sooooo much guys and i'll probably be on here asking for your help many more times. I hope you all don't get tired of me. Thanks a lot! :)

-K.

 

[kirsten_2009]

So here it is again...I think this time it's correct...

(1) Fe(s) + Cu2+(aq) --> Fe2+(aq) + Cu(s)

1.110 g Fe x (1 mol Fe / 56.0 g Fe) = 0.019821428 mol Fe

0.019821428 mol Fe x (1 mol Cu / 1 mol Fe) = 0.019821428 mol Cu

C = n/V ----> V = n/C
V= 0.019821428 mol Cu / 1.0M
V= 0.019821428
V = 0.020 L

I hope this is right! :) Again, thanks so much!

 

[Borek]

20 mL or 0.020 L it is.

Note: molar mass of iron is 55.845 g/mol. While it usually doesn't matter much and in most stoichiometry questions using 56 g/mol will be OK, stating it is 56.0 g/mol is incorrect. 55.8 g/mol if you want 3 siginificant digits.

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[kirsten_2009]

Thank you so much for all your help!

-K.

 

[kirsten_2009]

Hello! I have another question. Please let me know if it's o.k, and if not, could you point out what is wrong? Thanks!

Q: How many Cu (copper) atoms are present in a piece of sterlingsilver jewelry weighing 33.74g? (Sterling silver is a silver–copper alloy containing 92.5% Ag (silver) by mass.)


A: (My attempt at an answer) So...if 92.5% of the 33.74g sample of jewelry is made up of silver, 7.5% of the same piece of jewelry must be made of copper. So...

33.74 x (92.5 / 100) = 31.2

33.74 - 31.2 = 2.54

2.54 g Cu x (1 mol Cu / 63.546 g Cu) = 0.03997 mol Cu

0.03997 mol Cu x (6.022 x 10^23 / 1 mol Cu) = 2.41 x 10^22 atoms of Cu

Please let me know what you think ! :)

-K.

 

[Bohrok]

I did 33.74 g(.075) = 2.53 g and got 2.40×10²². Not much difference, but I think this way is more straightforward.

 

[kirsten_2009]

Hi! o.k sounds good! Thanks so much! :)

 

[Borek]

Remember to not round down intermediate results. If you show them - use rounded down, but do not use those rodunded down to calculate anything.

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[kirsten_2009]

Hello! Thanks, I will keep that in mind.

 

[kirsten_2009]

Hello everyone! I am back, as I had predicted. This time, I would really appreciate it if anyone could just go through these 3 questions and let me know if they think they're alright. Also, for the last part of the very last question, I have no idea how to solve it so if anyone can help me I would REALLY appreciate it, I am learning a lot here! :)

1.Q: // A solution contains 11.00% sucrose (cane sugar) by mass. What mass of the solution, in grams, is needed for an application that requires 2.80kg sucrose?

A: // I assumed a 100g sample of the sucrose solution.
2.80 kg sucrose x ( 1g / 1000kg ) = 0.0028g sucrose
0.0028g sucrose x (100g solution / 11.00g sucrose) = 0.0255g solution

2.Q: // Magnesium (Mg2+) occurs in seawater to the extent of 1.4g magnesium per kilogram of seawater. What volume of sea water, in cubic meters, would have to be processed to produce 1.30 x 10^5 ton of magnesium? Assume a density of 1.025g/mL for seawater. Remember, 1 ton = 2000 lb. Give your answer to two significant figures.

A: // 1.30 x 10^5 tons Mg2+ x (2000 lb / 1 ton) = 260000000 lb Mg2+
260000000 lb Mg2+ x (453.59237 g / 1 lb) = 1.18 x 10^11 g Mg2+
1.18 x 10^11 g Mg2+ (1 kg seawater / 1.4 g Mg2+) = 8.43 x 10^10 kg seawater
8.43 x 10^10 kg seawater (1 g / 1000 kg) = 84300000g
84300000g x (1 mL / 1.025g) = 82243902.44 mL
82243902.44 mL x (1m^3 / 1000000 mL) = 82 m^3

3.Q: // This question will be based on your knowledge of the law of multiple proportions.
i.)There are 2 different compounds of sulfur and fluorine. In SF_6, the mass of fluorine /g of sulfur is 3.55g, in the other compound , SF_x, the mass of fluorine /g of sulfur is 1.18g. What is the value of “x” for the second compound? (answer must be an integer).

ii.) Samples of 3 different compounds were analyzed and the masses of each element were determined as follows
Compound #1: Mass of Nitrogen 5.6g, Mass of Oxygen 3.2g
Compound #2: Mass of Nitrogen 3.5g, Mass of Oxygen 8.0g
Compound #3: Mass of Nitrogen 1.4g, Mass of Oxygen 4.0g

If you were John Dalton and had never heard of a mole, which of the following would you think were possible sets of formulas for the compounds #1,2 and 3, respectively? Chose all that apply

A.) NO_4, NO_10, NO_5
B.) NO, NO_2, NO_4
C.) N_2O, N_2O_4, N_2O_5
D.) NO_2, NO_8, NO_10

iii.)In one experiment, the burning of 0.318g sulfur produced 0.635g sulfur dioxide as the sole product of the reaction. In a second experiment, 0.844g sulfur dioxide was obtained. What mass of sulfur must have been burned in the second experiment?

A:// i.) SF_2 (But I think I guessed, I don’t know how to actually come up with the answer)
ii.) I chose “C” based on ratios I came up with...but don’t know if that’s the right
procedure.
iii.)This one, I have absolutely no idea how to solve it, if someone could help me I would most sincerely appreciate it!

 

[Borek]

It is 1000 g per 1 kg, not the other way around. k stands for kilo, that means thousand.

3rd can be done using ratios.

http://www.chembuddy.com/?left=balancing-stoichiometry&right=ratio-proportions

Note that you don't have to set up ratio using reaction equation. Take a look at "If you can buy 2 eggs for $1 how many eggs can you buy if you have $2?" Written as ratio it is

$$\frac {2_{eggs}} {1USD} = \frac {N_{eggs}} {2USD}$$

Solve for unknown number of eggs (N_eggs).

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methods

 

[kirsten_2009]

Hello Borek,

Thanks a lot for that, I corrected my mistake and I solved the third question and they were right. However, my professor (or rather her computer) said that question #2 is wrong...I can't find the mistake...is it because I didn't cube the centimetres in the last step? or is it because of another reason...I can't figure it out...if you spot anything could you let me know? Thanks again for everything!

-K.

 

[Borek]

First of all - question is ambiguous, as ton can be either a metric ton (1000 kg) or short ton (2000 lb) and it is not clearly stated which one it is. Could be that's local thing, but I am using metric in everydays life so for me that's 1000 kg, period.

Your problems started at 84300000g - that's not the correct mass. But I have already told you that.

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[kirsten_2009]

Hi Borek,

Again, thanks so much for your help in the last question it really helped me out. Nonetheless, I have another question and if you have a chance to help me out I'd appreciate it.

Q: 1.0560 g of antacid is weighed and mixed with 75.00mL of excess 0.1126 M HCL. The excess acid required 4.68mL of 0.1008 M NaOH for back titration. Calculate the neutralizing power of the antacid in terms of mmol H+ per gram of antacid.

A: So here is my approach. I figured I would figure out the moles of H in HCl in the following way 0.075 L x (0.1126 mol / 1 L) = 0.008445 mol HCl
0.008445 mol HCl x (1 mol H / 1 mol HCl) = 0.008445 mol H (since 1:1 ratio)...

Now I would use the balanced chemical equation and molar ratios to determine the moles of antacid that would react ...problem is though that I wouldn't know how to get a balanced chemical equation since I don't know the chemical formula for "antacid"...do you know of any other approach that I could use to solve this problem? Thanks a lot!

 

[Borek]

You don't have to know the formula for antacid, nor the reaction equation of antacid with hydrochloric acid. You know that some acid was in excess - can you calculate this amount from the information about amount of NaOH that was used for neutralization? using this information, can you calculate how much acid was neutralized by antacid?

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methods

 

[kirsten_2009]

Hello Borek,

So I took your advice and this is what I came up with, let me know what you think.

4.68mL x (1L/1000mL) = 0.00468 L
0.00468 L x (0.1008 mol / 1 L) = 0.000471744 mol NaOH

Therefore, 0.000471744 mol of NaOH were present in the second antacid and since it is a neutralization reaction the moles of NaOH must equal the moles of HCl so...
0.008445 mol HCl - 0.000471744 mol NaOH = 0.007973 mol HCl. Meaning that 0.007973 mols of HCl were neutralized by the first antacid...

0.007973 mol HCl = 0.007973 mol H+ because of 1:1 ratio.
0.007973 mol HCl x (1000 mmol / 1 mol HCl) = 7.9732 mmol HCl or mmol H+
7.9732 mol H+ / 1.0560 g antacid = 7.550 mmol H+/grams of antacid.

Do you think this is correct? Thank you sooooo much!

-K.

 

[Borek]

Wasn't that hard

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methods

 

[kirsten_2009]

Hi Borek! :)

Actually, it wasn't that hard as long as I was told in which direction to go. I think that's my biggest problem, I know the theory but I can't really problem solve. Thanks for your help!

I have another question though...and they probably won't stop so if at any moment you just feel like you've had enough of helping me no hard feelings :)

Acetylene torches are used for welding. These torches use a mixture of acetylene gas, , and oxygen gas, to produce the following combustion reaction:

Q: Imagine that you have a 5.00 gas tank and a 3.00 gas tank. You need to fill one tank with oxygen and the other with acetylene to use in conjunction with your welding torch. If you fill the larger tank with oxygen to a pressure of 115 , to what pressure should you fill the acetylene tank to ensure that you run out of each gas at the same time? Assume ideal behaviour for all gases.

A: I could figure out the moles of C2H2 I would need by figuring out the moles of O2 I have and using the balanced chemical equation to employ molar ratios:

If 1 mol of gas = 22.4 L
1 mol / 22.4 = 0.04464 mol
0.044646 mol x 5.00 L = 0.2232 mol of O2

0.2232 mol O2 x (2 mol C2H2 / 5 mol O2) = 0.08928 mol C2H2

So I thought I could use the following formula:

R = P1 V1 / n1 T1 = P2 V2 / n2 T2

I would cancel out T (temperature) since it’s not mentioned so I assume it’s kept constant and solve for P2...

P2 = 115atm x 5.00 L x 0.08928 mol C2H2 / 0.2232 mol O2 x 3.00 L

P2 = 76.7 atm.

This however seems counter-intuitive since if I lower the pressure shouldn’t the volume go up? What do you think?

-K.

 

[kirsten_2009]

Sorry...for the above post:

the combustion reaction is...

2C2H2 (g) + 5O2 (g) --> 4CO2 (g) + 2H2O (g)

The volume of the gas tanks is in Liters and the pressure is in atmospheres.

 

[Borek]

P2 = 76.7 atm.

This is the correct result, however

1 mol / 22.4 = 0.04464 mol

You have lost me here. If I understand correctly what you did, you first calculated number of moles in 1L of acetylene, and then used it to calculate number of moles of oxygene that will react with this 1L of acetylene. This is not necessary:

0.2232 mol O2 x (2 mol C2H2 / 5 mol O2) = 0.08928 mol C2H2

is equivalent to

2 mol C2H2 / 5 mol O2 = 0.08928 mol C2H2 / 0.2232 mol O2

so you don't have to use molar ratio to calculate molar ratio using 1L to finally use this molar ratio for 1L instead of stoichiometric molar ratio in the final equation :

P2 = 115atm x 5.00 L x 0.08928 mol C2H2 / 0.2232 mol O2 x 3.00 L

is equivalent to just

P2 = 115atm x 5.00 L x 2 mol C2H2 / 5 mol O2 x 3.00 L

Right?

Note, that actual pressure and volume units don't matter - as long as you use identical ones for both gases they simply cancel out!

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[kirsten_2009]

Hi Borek,

Ahhh...I see...it actually makes a lot of sense and it was easier. Thank you! As always!

 

[kirsten_2009]

Hi Borek,

Once more I have another question...this time it's about gas properties.

Q://Helium-oxygen mixtures are used by divers to avoid the bends and are used in medicine to treat some respiratory ailments. What percent (by moles) of He is present in a helium-oxygen mixture having a density of 0.528 g/L at 25 C and 721 mmHg? Express your answer in %He.

A:// I could use the ideal gas law PV=nRT to figure out the number of moles in the mixture but I have no idea how I could figure out the moles of just He. Any ideas? thanks in advance!

-K.

 

[kirsten_2009]

Hi Borek,

Nevermind...I got it :)

 

[Redbelly98]

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Redbelly98
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