Chem help!?

Molar mass of Ni₃(PO₄)₂ = (58.7×3 + 31.0×2 + 16.0×8) g/mol = 366.1 g/mol

2K₃PO₄(aq) + 3NiSO₄(aq) → Ni₃(PO₄)₂(s) + 3K₂SO₄(aq)
OR: Mole ratio K₃PO₄ : NiSO₄ = 2 : 3

Initial no. of moles of K₃PO₄ = (0.200 mol/L) × (100.0/1000 L) = 0.0200 mol
Initial no. of moles of NiSO₄ = (0.150 mol/L) × (200.0/1000 L) = 0.0300 mol

When K₃PO₄ completely reacted, NiSO₄ needed = (0.0200 mol) × (3/2) = 0.0300 mol
Hence, both the reactants completely react.

Refer to the above equation, mole ratio K₃PO₄ : Ni₃(PO₄)₂ = 2 : 1
No. of moles of K₃PO₄ reacted = 0.0200 mol
No. of moles of Ni₃(PO₄)₂ precipitate formed = (0.0200 mol) × (1/2) = 0.0100 mol
Mass of Ni₃(PO₄)₂ precipitate formed = (0.0100 mol) × (366.1 g/mol) = 3.66 g

It's kinda easy, you just need to write down the equation and do some simple calculations

2K3PO4 + 3NiSO4----> 3K2SO4+ (Ni)3(PO4)2

Molarity is basically no. of mole per litre of solution so for potassium phosphate, no. Of moles come out to be 0.2*0.1(L) =0.02 moles
Similarly for nickel(II) sulfate it comes out to be 0.15*0.2=0.03 moles
The precipitate here will be nickel phosphate
Using stoichiometry (no limiting reagent is present) 2 moles K3PO4 give 1 mole (Ni)3(PO4)2 so 0.02 moles would give 0.01 moles.
There you go!