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
