A student is working on a research project. The instructions in the book on how to prepare Ni(NH3)6Cl2 say that the percent yield in this preparation is 75.5 %. If the limiting reagent for the preparation is NiCl2·6H2O, and the student needs 50.0 grams of the Ni(NH3)6Cl2 product for the next step of the research project, how many grams of the NiCl2·6H2O should the student start with to prepare the desired quantity of the Ni(NH3)6Cl2.

%yield = (actual/theoretical)*100 = 75.5%

0.755 = (50/theoretical).
So you want to end up with 66.2g so that with 75.5% yield you will get 50.0 g. You can check that when you finish to see that you actually end up with 50.0 grams.

NiCl2.6H2O + 6NH3 ==> Ni(NH3)6Cl2 + 6H2O

mols Ni(NH3)6Cl2 = 66.2/molar mass = approx 0.29 but you should go through this and all of the other calculations to do them a bit more accurately.
Convert mols Ni(NH3)6Cl2 to mols NiCl2.6H2O. From the equation you see that is 1:1 so you have 0.29 mols starting material. How many grams is that? That's g = mols x molar mass = 0.29 x molar mass NiCl2.6H2O.

To find out how many grams of NiCl2·6H2O the student should start with, we need to consider the percent yield and the stoichiometry of the reaction.

1. Start by converting the desired mass of Ni(NH3)6Cl2 to moles. Since the molar mass of Ni(NH3)6Cl2 is not given, we can calculate it:
Ni(NH3)6Cl2 = Ni + 6(NH3) + 2(Cl)
From the periodic table, the molar mass of Ni is 58.69 g/mol, NH3 is 17.03 g/mol, and Cl is 35.45 g/mol. So the molar mass of Ni(NH3)6Cl2 is:
(58.69 g/mol) + 6(17.03 g/mol) + 2(35.45 g/mol) = 237.38 g/mol

Now, calculate the number of moles of Ni(NH3)6Cl2 needed:
Moles of Ni(NH3)6Cl2 = (Desired mass of Ni(NH3)6Cl2) / (Molar mass of Ni(NH3)6Cl2)
= (50.0 g) / (237.38 g/mol)

2. Next, we need to determine the stoichiometry of the reaction between NiCl2·6H2O and Ni(NH3)6Cl2. According to the balanced equation:
NiCl2·6H2O + 6 NH3 → Ni(NH3)6Cl2 + 6 H2O

It shows that 1 mole of NiCl2·6H2O reacts with 1 mole of Ni(NH3)6Cl2.

3. Based on the stoichiometry, the number of moles of NiCl2·6H2O required will be the same as the number of moles of Ni(NH3)6Cl2 needed. Therefore, the number of moles of NiCl2·6H2O needed is:
Moles of NiCl2·6H2O = Moles of Ni(NH3)6Cl2
Moles of NiCl2·6H2O = (50.0 g) / (237.38 g/mol)

4. Now, we need to consider the percent yield. If the percent yield is 75.5%, it means that only 75.5% of the theoretical yield is obtained. We can calculate the actual yield (in moles) by multiplying the desired yield (in moles) by the percent yield:
Actual yield (moles) = Percent yield * Theoretical yield (moles)

5. Finally, we can calculate the mass of NiCl2·6H2O needed, using the actual yield (moles) of NiCl2·6H2O and its molar mass:
Mass of NiCl2·6H2O = Actual yield (moles) * Molar mass of NiCl2·6H2O

By following these steps, you should be able to find the grams of NiCl2·6H2O that the student should start with to prepare the desired quantity of Ni(NH3)6Cl2.