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?

6.55 mL of K₃PO₄ are required

Explanation:

We need to propose the reaction, in order to begin:

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

Molarity = mol/L (Moles of solute that are contained in 1 L of solution.)

M = mol / volume (L). Let's find out the moles of chloride:

- We first convert the volume from mL to L → 187 mL. 1L / 1000mL = 0.187L

0.0184 M. 0.187L = 0.00344 moles of NiCl₂

Ratio is 3:2. Let's propose this rule of three:

3 moles of chloride react with 2 moles of phosphate

Then, 0.00344 moles of NiCl₂ will react with (0.00344. 2) / 3 = 0.00229 moles of K₃PO₄

M = mol / volume (L) → Volume (L) = mol/M

Volume (L) = 0.00229 mol / 0.350 M = 6.55*10⁻³L

We convert the volume from L to mL → 6.55*10⁻³L. 1000mL / 1L = 6.55 mL