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18 May, 19:22

What volume of 0.205 m k3po4 solution is necessary to completely react with 154 ml of 0.0110 m nicl2? 1.65 l?

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  1. 18 May, 21:14
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    The balanced equation for the above reaction is

    2K₃PO₄ + 3NiCl₂ - - - > 6KCl + Ni₃ (PO₄) ₂

    stoichiometry of K₃PO₄ to NiCl₂ is 2:3

    the number of NiCl₂ moles reacted - 0.0110 mol/L x 0.154 L = 1.69 x 10⁻³ mol

    if 3 mol of NiCl₂ reacts with - 2 mol of K₃PO₄

    then 1.69 x 10⁻³ mol of NiCl₂ reacts with - 2/3 x 1.69 x 10⁻³ = 1.13 x 10⁻³ mol of K₃PO₄

    molarity of K₃PO₄ solution given - 0.205 M

    there are 0.205 mol in 1 L

    therefore 1.13 x 10⁻³ mol are in - 1.13 x 10⁻³ mol / 0.205 mol/L = 5.51 mL

    volume of K₃PO₄ required - 5.51 mL
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