calculate the theorectical yield of the compounds to be prepared in the experiment. The metal ion in both cases is the limiting reagent.Find nuber of moles of Cu(II) in the sample of CuSO4 x 5H2O that you used. The equal the number of moles of [Cu(NH3)4]x H2O that could theoretically be prepared. Proceed in a similar way for the synthesis invloving Co(II).

This what I got.

CuSO4 x 5H20
mass=3.86g
3.86/249.70g/mol=1.526x10-4 CuSO4
each mole CuSO4x5H20 will dissociate in 1 mole Cu2+ and 1 mol SO4 2-, giving a total of 2 moles of ions.



Also, how would i establish that the metal ion is the limiting reagent?

You have posted this a number of times and I see you haven't received any help. I've resisted because the post isn't clear to me.If you will go through and type in the missing words (two or three of them are missing or they are typos) I might be able to make sense of the question. My quick answer about limiting reagent is that I see no mention of amount of NH3. Without that someone must TELL you which is the limiting reagent; the problem does that.

Here is a link that has nothing to do with copper ammine complex; however, it will show you how to calculate theoretical yield and percent yield.
http://www.jiskha.com/science/chemistry/stoichiometry.html

I am also facing this problem. The question is correctly asked. The amount NH3 is not given, but the problem states that the metal ion is the limiting reagent.

To calculate the theoretical yield of the compounds and determine the limiting reagent, we need to use the stoichiometry of the reactions.

1. Calculate the moles of Cu(II) in the sample of CuSO4 x 5H2O:
moles CuSO4 = mass CuSO4 / molar mass CuSO4
moles CuSO4 = 3.86 g / 249.70 g/mol
moles CuSO4 = 0.01543 mol CuSO4

2. According to the equation CuSO4 x 5H2O → [Cu(NH3)4]x H2O, the ratio is 1:1. This means that for every 1 mole of CuSO4 used, 1 mole of [Cu(NH3)4]x H2O is formed.

Therefore, the number of moles of [Cu(NH3)4]x H2O that could theoretically be prepared is also 0.01543 mol.

For the synthesis involving Co(II), the same steps can be followed. You need to find the moles of Co(II) and use the stoichiometry of the reaction to determine the theoretical yield.

To establish that the metal ion is the limiting reagent, you need to compare the moles of the metal ion (Cu(II) or Co(II)) to the moles of the other reactant (ammonia in both cases). The reactant with fewer moles is the limiting reagent because it will be completely consumed in the reaction, limiting the amount of product formed.

For example, if you have 0.01543 mol of Cu(II) and 0.025 mol of ammonia, Cu(II) is the limiting reagent because you have less of it compared to ammonia.

To calculate the theoretical yield of the compounds to be prepared in the experiment, you need to determine the number of moles of Cu(II) in the sample of CuSO4 x 5H2O that you used.

First, find the mass of CuSO4 x 5H2O that you have:
mass = 3.86g

Then, calculate the number of moles of CuSO4 x 5H2O:
moles = mass / molar mass
molar mass of CuSO4 x 5H2O = 249.70 g/mol
moles = 3.86g / 249.70 g/mol
moles = 1.526x10^-4 moles of CuSO4 x 5H2O

Since each mole of CuSO4 x 5H2O will dissociate into 1 mole of Cu2+ ions, you have an equal number of moles of Cu2+ ions.

Now, let's proceed with the synthesis involving Co(II). Using a similar approach:

Find the mass of the compound involved in the synthesis.
Calculate the number of moles of the compound.
Determine the number of moles of Co(II) that could theoretically be prepared.

To establish that the metal ion is the limiting reagent, you need to compare the stoichiometric ratio of the metal ion to the other reactants.

For example, if you have the reaction:
2Cu2+ + 3NH3 → [Cu(NH3)4]2+

You need to compare the mole ratio between Cu2+ and NH3. If, for example, you have 1 mole of Cu2+ ions and only 2 moles of NH3 available, then Cu2+ is the limiting reagent because you don't have enough NH3 to react with all the Cu2+ ions. The limiting reagent determines the maximum amount of product that can be formed.