Question

How many liters of 0.305 m K3PO4 solution are necessary to completely react with 187 ml of 0.0184M NiCl2 according to the balanced chemical reaction:

2 K3PO4(aq) + 3NiCl2(aq)>Ni3(PO4)2(s) +6 KCl(aq)

Answer 1

0.00752L of 0.305M K₃PO₄

Step-by-step explanation:

Based on the balanced chemical equation, 2 moles of K₃PO₄ react with 3 moles of NiCl₂.

To find the volume of K₃PO₄ we need to determine the moles of this compound required for a complete reaction finding moles of NiCl₂ and using the chemical equation:

Moles NiCl₂:

187mL = 0.187L * (0.0184mol / L) = 0.00344 moles NiCl₂

Moles K₃PO₄:

As 2 moles of K₃PO₄ react with 3 moles of NiCl₂:

0.00344 moles NiCl₂ * (2 mol K₃PO₄ / 3 moles NiCl₂) =

0.002294 moles of K₃PO₄

Volume of 0.305M K₃PO₄:

0.002294 moles of K₃PO₄ * (1L / 0.305moles K₃PO₄) =

0.00752L of 0.305M K₃PO₄