Question

1.What is the maximum amount of  [Ni(NH₃)₆]Cl₂ product (the theoretical yield), that can be made if you start with 5.107 g of NiCl₂·6H₂O?

Data:
[Ni(NH₃)₆]Cl₂ = 231.77 g/mol
NiCl₂·6H₂O = 237.71 g/mol

2.For synthesis experiments, purification steps tend to

a. decrease yield

b. can't be generalized

c. not affect yield

d. increase yield

3. What is the %yield for a reaction that produces 4.781 g of [Ni(NH₃)₆]Cl₂ if the theoretical yield is 5.060 g?

Answer 1

NiCl2.6H2O + 6NH3 ----> Ni(NH3)6Cl2 + 6H2O
Based on this reaction lets calculate moles of products
here mole ratio is 1:1

5.107 g NiCl2.6H2O= [1 mole/237.71 g/mol(molar mass)]*5.107g
=0.0215 moles NiCl2.6H2O ***

since mole ratio is 1:1
0.0215 moles NiCl2.6H2O produce 0.0215 mols of [Ni(NH3)6]Cl2

moles of [Ni(NH3)6]Cl2 = 0.0215

mass of [Ni(NH3)6]Cl2
= moles of [Ni(NH3)6]Cl2 x molar mass of [Ni(NH3)6]Cl2
= 0.0215 x 231.77 g/mol = 4.98 g***
Thus theoretical yeild of [Ni(NH3)6]Cl2= 4.98 g
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2. Purification process tend to decrease yield***
- it can be due to our experimental errors or impurities present etc
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3. % yield= Actual yield/theoritical yield *100 =
= 4.781 g/5.06 g *100 =94.48 %
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Thank you :)