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 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 establishthat the metal ion is the limiing reagent?

To calculate the theoretical yield of the compounds to be prepared in the experiment, you can follow these steps:

1. Determine the number of moles of Cu(II) in the sample of CuSO4 · 5H2O that you used:
- Given that the mass of CuSO4 · 5H2O is 3.86g and the molar mass is 249.70g/mol.
- Divide the mass by the molar mass: 3.86g / 249.70g/mol = 0.0154 mol CuSO4 · 5H2O.

2. Each mole of CuSO4 · 5H2O will dissociate to give one mole of Cu2+ ions, so the number of moles of Cu(II) is also 0.0154 mol.

3. To calculate the moles of [Cu(NH3)4] · H2O that could theoretically be prepared, you need to use stoichiometry. The reaction is:
CuSO4 · 5H2O + 4NH3 -> [Cu(NH3)4] · H2O + H2SO4

4. The stoichiometric ratio tells you that one mole of CuSO4 · 5H2O reacts with four moles of NH3 to produce one mole of [Cu(NH3)4] · H2O.

5. Therefore, the number of moles of [Cu(NH3)4] · H2O that could theoretically be prepared is also 0.0154 moles.

6. Repeat the same steps for the synthesis involving Co(II), using the appropriate masses and molar masses of the reactants and products.

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+) with the stoichiometric ratio in the reaction.

In this case, the stoichiometric ratio tells you that one mole of CuSO4 · 5H2O reacts with one mole of Cu2+ ions. If the number of moles of Cu2+ ions is equal to or less than the number of moles of CuSO4 · 5H2O, then the metal ion is not the limiting reagent.

However, if the number of moles of Cu2+ ions is greater than the number of moles of CuSO4 · 5H2O, then the metal ion (Cu2+ or Co2+) is the limiting reagent because it determines the maximum amount of the product that can be formed.

To calculate the theoretical yield of the compounds, we need to determine the number of moles of Cu(II) and Co(II) present in the starting compounds, CuSO4 x 5H2O and CoCl2 x 6H2O, respectively.

1. CuSO4 x 5H2O:
- The molar mass of CuSO4 x 5H2O is 249.70 g/mol.
- The mass of the sample used is 3.86 g.
- Divide the mass by the molar mass to find the number of moles:
Moles of CuSO4 x 5H2O = 3.86 g / 249.70 g/mol = 0.01546 mol.

- In CuSO4 x 5H2O, each mole dissociates into 1 mole of Cu2+ ions and 1 mole of SO4 2- ions, giving a total of 2 moles of ions. Therefore:
Moles of Cu2+ ions = 2 x 0.01546 mol = 0.03092 mol.

- The number of moles of [Cu(NH3)4]x H2O that could theoretically be prepared is equal to the number of moles of Cu2+ ions, as the formation of [Cu(NH3)4]x H2O requires 1 mole of Cu2+ ions.
Moles of [Cu(NH3)4]x H2O = 0.03092 mol.

2. CoCl2 x 6H2O:
- The molar mass of CoCl2 x 6H2O is 237.93 g/mol.
- The mass of the sample used is 1.78 g.
- Divide the mass by the molar mass to find the number of moles:
Moles of CoCl2 x 6H2O = 1.78 g / 237.93 g/mol = 0.007487 mol.

- In CoCl2 x 6H2O, each mole dissociates into 1 mole of Co2+ ions and 2 moles of Cl- ions, giving a total of 3 moles of ions. Therefore:
Moles of Co2+ ions = 3 x 0.007487 mol = 0.022461 mol.

- The number of moles of [Co(NH3)6]Cl3 that could theoretically be prepared is equal to the number of moles of Co2+ ions, as the formation of [Co(NH3)6]Cl3 requires 1 mole of Co2+ ions.
Moles of [Co(NH3)6]Cl3 = 0.022461 mol.

To establish whether the metal ion is the limiting reagent, we compare the moles of the metal ion (Cu2+ or Co2+) to the moles of the ligand (NH3). The metal ion is considered the limiting reagent if the moles of the metal ion are lesser than the moles of the ligand, indicating that it will run out first and limit the amount of product that can be formed.