If 3 moles of CuSO4 is needed to oxidize Fe to +3, how many mL of 1M CuSO4 do i need?

Question

i keep on thinking 3000, but considering this is a lab for school it seems reeeeeeeeeeeeeeeeeeeally unreasonable.

its for 1 g of fe

Answer 1

You would indeed need 3000 mL, or 3L, of 1M solution to have 3 moles of CuSO4. However, you don't need three moles. You only need three times the number of moles of iron you have, because 3 moles of CuSO4 oxidizes 1 mole of Fe. Fe has a molar mass of 55.845 g, so you have just 1 / 55.845 = 0.0179 moles of iron, and you only need 3 * 0.0179 = 0.0537 moles of CuSO4. With your 1M solution, you only need 53.7 mL of solution, which looks a lot more manageable. If 3 moles of CuSO4 actually neutralizes 2 moles of Fe, as suggested below, then you'd need only half the amount I calculated, or 26.9 mL.

Answer 2

First, find out how much 1g of Fe is in moles. I always use this proportion:

55.84g of Fe 1 g of Fe x = 1 mole of Fe * 1 g of Fe * 1
------------------ = ---------------------- -----------
1mole of Fe x mole of Fe 55.84 g of Fe


Doing the math, you get 0.0179 moles of Fe.

Now, let's write the equation for the oxidation of Fe by Cu:

3Cu+ + Fe ---> 3Cu + Fe3+

According to this: 3 moles of Cu+ is used to oxidize 1 mole of Fe.

Therefore, the ratio of Cu+ and Fe here is 3:1

So, if we have 0.0179 moles of Fe, then the amount of Cu+ needed is 3 * 0.0179 moles = 0.0537 moles.

Molarity = moles of Cu+/volume(in liters); so:

1M = 0.0537 moles/x liters x = 0.0537 liters

To find out how much it is in mL, just multiply the above by 1000, since there are 1000 mLs in a liter.

Therefore, our final answer is: 53.7 mL

Hope this helps, and I hope that you practice more of these questions. After all, "practice makes perfect".

Answer 3

The relation between Moles, Morality (concentration) and volume (in L or dm3) is as follows: Conc.= Moles/Volume

Changing form... Volume=moles/ conc.
So V=3/1
= 3 dm3 (or 3L)

Now I dm3 is 1000 cm3 (mL) so you do in fact need 3000 mL... Don't worry as long as you did the initial calculations correctly you shouldn't worry about the practicality of the assignment.

Answer 4

I don't have an inorganic chemistry text to verify this but I found on the web site below that its:

Cu++ and Fe -----------> Cu and Fe++

maybe it can go farther to oxidize it to Fe+++, I don't know.

But based on this equation the following is what I calculate:

(1g Fe)*(1mole Fe/55.847 g)*(1 mole Cu++/1 mole Fe) = 0.0179 moles of Cu++ needed.

(0.0179 mole Cu++) = (1mole Cu++/1000ml)*(x ml),
solve for x = 17.9 ml

http://genchem.chem.wisc.edu/sstutorial/Text14/Tx141/tx141.html