What volume of 0.344 M K3PO4 is required to
react with 79 mL of 0.87 M MgCl2 according
to the equation
2 K3PO4 + 3 MgCl2 → Mg3(PO4)2 + 6 KCl
Answer in units of mL.
79mL * 0.87moles/L = 0.06873 moles of MgCl2
So, that will react with 2/3 that many moles of K3PO4: 0.04582 moles
.04582 moles * 1L/0.344moles = 0.133L = 133 mL