Question

Consider the reaction:
2K3PO4(aq)+3NiCl2(aq)→ Ni3(PO4)2(s)+6KCl(aq)

What volume of 0.205 M K3PO4 solution is necessary to completely react with 114 mL of 0.0110 M NiCl2?

Answer needed in L

Answer 1

Number of moles of NiCl2 = molarity * volume of solution in L

Number of moles of NiCl2 = 0.0110 * 0.114 L = 0.001254 mole

From the balanced equation we can say that

3 mole of NiCl2 requires 2 mole of K3PO4 so

0.001254 mole of NiCl2 will require

= 0.001254 mole of NiCl2 *( 2 mole of K3PO4 / 3 mole of NiCl2)

= 0.000836 mole of K3PO4

Number of moles of K3PO4 = 0.000836

Molarity = number of moles / volume of solution in L

0.205 = 0.000836 / volume of solution in L

volume of solution in L = 0.000836 / 0.205 = 0.00408 L

Therefore, the volume of K3PO4 required would be 0.00408 L