I've been stuck on this equation for 2 hours, please help?

No. of moles of KMnO₄ used = (0.020 mole/L) × (25/1000 L) = 0.0005 mole

1 mole of KMnO₄ contains 1 mole of MnO₄⁻ ions.
No. of moles of MnO₄⁻ ions reacted = 0.0005 mole

Net ionic equation for the reaction :
5Fe²⁺(aq) + MnO₄⁻(aq) + 8H⁺(aq) → 5Fe³⁺(aq) + Mn²⁺(aq) + 4H₂O(l)
Mole ratio Fe²⁺ : MnO₄⁻ = 5 : 1
No. of moles of Fe²⁺ ions required = (0.0005 mole) × 5 = 0.0025 mole

1 mole of Fe(NH₄)₂(SO₄)₂•6H₂O contains 1 moles of Fe²⁺ ions.
No. of moles of Fe(NH₄)₂(SO₄)₂•6H₂O required = 0.0025 mole

Molar mass of Fe(NH₄)₂(SO₄)₂•6H₂O = (55.8 + 14.0×2 + 1.0×20 + 32.1×2 + 16.0×14) g/mol = 392.0 g/mol
Mass of Fe(NH₄)₂(SO₄)₂•6H₂O required = (0.0025 mole) × (392.0 g/mol) = 0.98 g

Potassium permanganate reacts with iron salts as shown below.

5 X (Fe2+ Fe3+ + e-) oxidation

MnO4- + 8H+ + 5e- Mn2+ + 4H2O reduction

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MnO4- + 5Fe2+ + 8H+ Mn2+ + 5Fe3+ + 4H2O overall process

Therefore for every 1 mole of MnO4-, 5 moles of Fe2 are needed.

We determine the amount of moles of MnO4- in solution by multiplying 0.02 M (mol/L) by 0.025 L (25mL). This gives 0.0005 moles of MnO4-. We multiply this by 5 to determine the moles of Fe2 needed. This gives us 0.0025 moles of Fe2. Fe2 and Fe(NH4)2(SO4)2 * 6H2O has a 1 to 1 molar ratio, so we can say 0.0025 moles of Fe(NH4)2(SO4)2 * 6H2O are needed. We multiply this by the molar mass of Fe(NH4)2(SO4)2 * 6H2O (392.14g/mol). This gives us 0.98035 g of Fe(NH4)2(SO4)2 * 6H2O needed to react with 25mL of 0.02M KMnO4.