What volume of 0.408 M K3PO4 is required to react with 28 mL of 0.75 M MgCl2 according to the equation?

2 K₃PO₄ + 3 MgCl₂ → Mg₃(PO₄)₂ + 6 KCl
OR: Mole ratio K₃PO₄ : MgCl₂ = 2 : 3

No. of milli-moles of MgCl₂ = (0.75 mmol/mL) × (28 mL) = 21 mmol
No. of milli-moes of K₃PO₄ = (21 mmol) × (2/3) = 14 mmol
Volume of K₃PO₄ = (14 mmol) / (0.408 mmol/mL) = 34.3 mL

millimoles Mg(Cl)2 ---> (0.75 mmol/mL) (28 mL) = 21 mmol

K3PO4 and MgCl2 react in a 2:3 molar ratio

2 is to 3 as x is to 21

x = 14 mmol (of K3PO4 required)

0.408 mmol/mL = 14 mmol / y

y = 34.3 mL (round off to 34 if so desired)

no idea