Respuesta :
Answer:
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
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,
[tex]\frac{2}{3} \times 0.0034408[/tex] moles of K₃PO₄ will be required to completely react with 0.003408 moles of NiCl₂
[tex]\frac{2}{3} \times 0.0034408 = 0.00229387[/tex]
∴ 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
[tex]Volume = \frac{Number\ of\ moles}{Concentration}[/tex]
From the question
Concentration of K₃PO₄ = 0.305 M
∴ Volume of K₃PO₄ that are necessary = [tex]\frac{0.00229387}{0.305}[/tex]
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
Learn more here: https://brainly.com/question/15529766
