%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.
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.
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