[hongkong]

If 4.97 mL of 0.16 M copper(II) sulfate and 8.25 mL of  2.8 M aqueous ammonia are mixed and diluted to 44.73 mL, what will be the copper complex concentration in this solution?

The answer is 0.018, but I got like 0.129. I don't know how to get 0.018. Can you guys please help me! Thanks a lot!

[Wald_ron]

The first step is to write the equation and balance it.

hint:
aq ammonia= NH₄OH

[FreeTheBee]

How much copper will there be present in the final solution and how much ammonia? And how did you get to your answer?

[hongkong]

I try, but it's not the  right answer

CuSO4 + 4NH3 ---> Cu(NH3)4SO4

4.97 ml=0.00497L
8.25ml= 0.00825L

mole of Cu2SO4 = 0.00497L x 0.16mol/L = 7.952 x 10^-4
mole of NH3 = 0.00825L x 2.8mol/L= 0.0231mol

Then I find limiting reagent:
7.952 x 10^-4 mol CuSO4 x (4mol NH3/ 1molCuSO4) = 0.0032 mol NH3

0.0231molNH3  x (1molCuSO4/4 mol NH3)=0.0058 molCuSO4

Limiting reagent is CuSO4.

Cu + SO4 ----->CuSO4         44.73 mL= 0.04473L     
0.0058 molCuSO4/ 0.04473L = 0.129 M Cu

[Wald_ron]

I try, but it's not the  right answer

CuSO4 + 4NH3 ---> Cu(NH3)4SO4

Your formula is wrong

CuSO₄ + NH₄OH  (NH₄)₂SO₄ + Cu(OH)₂

is the unbalanced equation.

[hongkong]

But, is the method that I am doing is right or not 

Thanks 

[Wald_ron]

But, is the method that I am doing is right or not 

Thanks 

It's not right, the chemical formula is wrong, and you did the mathematics wrong, to determine the limiting reagent you have to figure out how much of each reagent produces the desired product.

for examples a + b  c + d

mols a * (mols c/ mols a) = mols c

mols b * (mols c/mols b) = mols c

and the smaller is the limiting reagent.

I hope this helps 

[hongkong]

Actually, what I did is all right except for the limiting reagent  . CuSO4 is limiting reagent, but I used NH3 for that  . So it was wrong. Now I got my answer Thanks you all!