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?

The volume of K₃PO₄ that are necessary to completely react with the NiCl₂ is 7.52 mL

From the question,

We are to determine the volume of K₃PO₄ necessary to completely neutralize the given NiCl₂

First, we will write a balanced chemical equation for the reaction

The balanced chemical equation for the reaction is

2K₃PO₄ + 3NiCl₂ → Ni₃(PO₄)₂ + 6KCl

This means

2 moles of K₃PO₄ are required to completely react with 3 moles of NiCl₂ to produce 1 mole of Ni₃(PO₄)₂ and 6 moles KCl

Now, we will determine the number of moles of NiCl₂ present

From the question

Concentration of NiCl₂ = 0.0184 M

Volume of NiCl₂ = 187 mL = 0.187 L

Using the formula

Number of moles = Concentration × Volume

∴ Number of moles of NiCl₂ present = 0.0184 × 0.187

Number of moles of NiCl₂ present = 0.0034408 moles

From the balanced chemical equation

2 moles of K₃PO₄ are required to completely react with 3 moles of NiCl₂

Then,

moles of K₃PO₄ will be required to completely react with 0.003408 moles of NiCl₂

∴ 0.00229387 moles of K₃PO₄ will be required to completely react with the 0.003408 moles of NiCl₂

Now, we will determine the volume of K₃PO₄ that will be required

From the formula

From the question

Concentration of K₃PO₄ = 0.305 M

∴ Volume of K₃PO₄ that are necessary =

Volume of K₃PO₄ that are necessary = 0.00752088 L

Volume of K₃PO₄ that are necessary = 7.52088 mL

Volume of K₃PO₄ that are necessary ≅ 7.52 mL

Hence, the volume of K₃PO₄ that are necessary to completely react with the NiCl₂ is 7.52 mL

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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