I need the whole equation layout not just the final result. I need a complete equation with units please. **Chemical Reaction Calculation****Problem Statement:**How many liters of a 0.305 M K₃PO₄ solution are necessary to completely react with 187 mL of a 0.0184 M NiCl₂ solution, according to the balanced chemical reaction:\[ 2 \text{ K₃PO₄(aq)} + 3 \text{ NiCl₂(aq)} \rightarrow \text{Ni₃(PO₄)₂(s)} + 6 \text{ KCl(aq)} \]**Calculation Tool:**- **Starting Amount:** - Two input boxes are available to enter starting amounts.- **Conversion Factors:** - A matrix of conversion factors is provided. - Options available include: - Constants: 1, 6, 0.001, 0.305, 0.00376, \( 6.022 \times 10^{23} \), 0.0184, 7.52, 0.0113, 0.00752. - Molarity and volume options: M NiCl₂, mol NiCl₂, g NiCl₂, mol K₃PO₄, L NiCl₂, M K₃PO₄, mL K₃PO₄, g K₃PO₄, mL NiCl₂, L K₃PO₄. - **Calculation Process:** - Select conversion factors and define relationships using the "ADD FACTOR" section. - Compute your answer using the "ANSWER" box. - Reset selections using the "RESET" button to start over if necessary.This interactive tool guides you through complex stoichiometric conversions and calculations, ensuring accurate chemical balancing.

Balanced reaction : 2K3PO4 + 3NiCl2 ....> Ni3(PO4)2 + 6KCl Molarity of NiCl2 = 0.0184 M Volume = 187 ml = 187ml×1L/1000ml = 0.187L Molarity = moles ÷ volume Moles of NiCl2 = (0.0184mol/L)×0.187L = 0.003441 moles By reaction stochiometry: (2mol K3PO4/3mol NiCl2)×0.003441 mol NiCl2 = 0.002294 mol K3PO4 Molarity of K3PO4 = 0.305M Volume of K3PO4 = 0.002294mol ÷ 0.305mol/L = 0.00752 L = 0.0075L×1000ml/1L = 7.52ml So, 7.52 ml of K3PO4 is needed.