Need help.... calculate the theorectical yield of the compounds to be prepared in the experiment. The metal ion in both cases is the limiting reagent.Find number 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.
1.546x10-2 mols CuSO4 x 2= 3.09 x 10-2 mole ions

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

How is the moles changeing? You have 1.526E-4 moles of Cu+2 ions, which will prepare then 1.526E-4 moles of the product. Why areyou changin numbers? What copper you have on one side is = to the copper on the other.

To calculate the number of moles of Cu(II) in the sample of CuSO4 x 5H2O, you correctly divided the mass of CuSO4 x 5H2O by its molar mass. Now to find the number of moles of Cu(II), you need to consider its stoichiometry with CuSO4 x 5H2O.

From the equation CuSO4 x 5H2O → Cu(II) + SO4(II), you can see that one mole of CuSO4 x 5H2O dissociates to form one mole of Cu(II) ions. Therefore, the number of moles of Cu(II) is the same as the number of moles of CuSO4 x 5H2O.

In this case, you calculated the number of moles of CuSO4 x 5H2O as 1.526x10-4 mol, so that is also the number of moles of Cu(II) ions.

Now, to determine if the metal ion is the limiting reagent, you need to compare the number of moles of Cu(II) ions to the number of moles of Cu(II) ions that could theoretically be prepared from [Cu(NH3)4]x H2O. The equation for this synthesis reaction may be given to you.

Let's say the balanced equation for the synthesis of [Cu(NH3)4]x H2O from Cu(II) is:

Cu(II) + 4NH3 → [Cu(NH3)4]x H2O

From the equation, you can see that one mole of Cu(II) reacts with four moles of NH3 to form one mole of [Cu(NH3)4]x H2O. Therefore, the number of moles of [Cu(NH3)4]x H2O that can theoretically be prepared is four times the number of moles of Cu(II).

Using the number of moles of Cu(II) you calculated earlier (1.526x10-4 mol), you can now determine the theoretical yield of [Cu(NH3)4]x H2O. Multiply the number of moles of Cu(II) by four:

1.526x10-4 mol Cu(II) x 4 = 6.104x10-4 mol [Cu(NH3)4]x H2O

This is the theoretical yield of [Cu(NH3)4]x H2O that could be prepared from the given amount of CuSO4 x 5H2O.

Similarly, you can proceed with the synthesis involving Co(II) by following the same steps and using the relevant stoichiometry.

To establish that the metal ion is the limiting reagent, you need to compare the number of moles of the metal ion (Cu2+ or Co2+) in the starting compound (CuSO4 x 5H2O or CoCl2 x 6H2O) with the number of moles of the metal ion required to form the complex compound ([Cu(NH3)4]x H2O or [Co(NH3)6]Cl2).

In the case of CuSO4 x 5H2O, you calculated that you have 3.09 x 10^-2 moles of ions (Cu2+ and SO4 2-). To determine if Cu2+ is the limiting reagent, you need to compare this with the number of moles of Cu2+ required to form the complex compound.

For example, if the balanced equation for the reaction is:
Cu2+ + 4NH3 -> [Cu(NH3)4]2+

You would need 1 mole of Cu2+ to react with 4 moles of NH3 to form 1 mole of [Cu(NH3)4]2+. Therefore, compare the number of moles of Cu2+ (3.09 x 10^-2 moles) with the number of moles required to form the complex compound. If the number of moles of Cu2+ is greater, then Cu2+ is not the limiting reagent and some excess Cu2+ will remain after the reaction. Otherwise, if the number of moles of Cu2+ is less, then Cu2+ is the limiting reagent and you would be able to form the maximum possible moles of [Cu(NH3)4]2+.

You can proceed similarly for the synthesis involving Co(II), comparing the number of moles of Co2+ with the number of moles required to form the complex compound.