[lifepi]

A solution is prepared by dissolving 25.0g of ammonium sulfate in enough water to make 100.0mL of stock solution. A 10.00mL sample of this stock solution is added to 50.00mL of water. Calculate the concentration of ammonium ions and sulfate ions in the final solutions.

My attempt:
25g (NH₄)₂SO₄ x (1mol (NH₄)₂SO₄/132.2g) = .19mol (NH₄)₂SO₄

(NH₄)₂SO₄ --> 2NH₄⁺ + SO₄^-2 ; NH₄⁺ = 2 x .19 = .38mol NH₄⁺

SO₄^-2 = 1 x (.19) = .19mol SO₄^-2

So I found the moles of ammonium sulfate and multiplied it by the number of moles of Ammonium (2) in the equation. And did the same with Sulfate.
Help with steps please
Thanks so much!

[Borek]

So far so good.

What is concentration of both ions in the first solution?

How much ammonium sulfate is transferred to the other solution?

What are the concentrations in the final solution?

[lifepi]

The concentration would be: M = m/L; .19mol (NH4)₂SO4 / .100L = 1.9M

The amount of Sulfate transferred would be: .19mol SO₄ x (96.07g / 1mol SO₄) = 18.25mol SO₄

The concentration in the final solution would be: 

[AWK]

Hint - dilution

[Borek]

The amount of Sulfate transferred would be: .19mol SO₄ x (96.07g / 1mol SO₄) = 18.25mol SO₄

No.  You have calculated mass of 0.19 mole of SO₄ (whatever it means), not number of moles of substance transefered.

How much sulfate in 10 mL of 1.9M solution?