Question

What volume of 0.205 mol/L K3PO4 solution is necessary to completely react with 126 mL of 0.0104 mol/L NiCl2?

Answer 1

$2K_3PO_4 + 3NiCl_2= Ni_3(PO_4)_2 + 6KCl$

$moles \space of \space NiCl_2 = V \cdot M= 0.126 L \cdot 0.0104 \frac {mol}{L} = 0.0013104 \space moles$

$moles of K_3PO_4 = \frac {0.0013104 \cdot 2}{3}= 0.0008736 \space moles$

$Volume \space of \space K_3PO_4 = \frac {n}{M}= \frac {0.0008736 moles}{0.205 \frac {moles}{L}}= 0.00426 L = 4.26 mL$