How many liters of 0.305 M K₃PO₄ solution are necessary to completely react

Question

How many liters of 0.305 M K₃PO₄ solution are necessary to completely react with 187 mL of 0.0184 M NiCl₂ according to the balanced chemical reaction: 2 K₃PO₄(aq) + 3 NiCl₂(aq) → Ni₃(PO₄)₂(s) + 6 KCl(aq)

Answer 1

Start with given 187 ml = 0.187 L of NiCl₂ and its Molarity 0.0184 mole/L. Multiply together to get 0.00344 moles of NiCl₂. From the balanced reaction, (0.00344 mole NiCl₂) * (2 mole K₃PO₄ / 3 mole NiCl₂) = 0.00229₄ mole K₃PO₄ [I keep an insignificant digit often to avoid round-off error]. Divide this by the desired Molarity: (0.00229₄ mole K₃PO₄ / (0.305 mole/L) = 0.00752 L of K₃PO₄ or 7.52 mL.