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Need help.... calculate the theorectical yield of the compounds to be prepared in the experiment. The metal ion in both cases i...Asked by Debbie
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?
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?
Answers
Answered by
bobpursley
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.
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