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.
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?
To calculate the theoretical yield of the compounds, you first need to determine the number of moles of Cu(II) in the sample of CuSO4 x 5H2O used.
In the given information, you have the mass of CuSO4 x 5H2O as 3.86g. To find the number of moles, divide the mass by the molar mass of CuSO4 x 5H2O, which is 249.70g/mol.
So, moles of CuSO4 x 5H2O = 3.86g / 249.70g/mol = 0.01526 mol
Since each mole of CuSO4 x 5H2O dissociates into 1 mole of Cu2+, you have 0.01526 mol of Cu2+ ions.
Now, for the synthesis involving Co(II), you need to repeat the same steps, but using the appropriate values. Let's assume you have the mass of CoCl2 x 6H2O. After calculating the moles of Co(II), you will have the number of moles of [Co(NH3)4]x H2O that can theoretically be prepared.
To determine if the metal ion is the limiting reagent, you need to compare the moles of the metal ion (Cu2+ or Co(II)) with the moles of the other reactant (NH3 in the case of [Cu(NH3)4]x H2O and NH3 + H2O in the case of [Co(NH3)4]x H2O). Whichever reactant has the lower ratio compared to its stoichiometric coefficient would be the limiting reagent. In other words, the reactant with fewer moles compared to the balanced equation is the limiting reagent.