Question

1. Balance the reaction below and determine how many grams of nickel(II) phosphate will be formed when sodium phosphate (12.05g) is mixed with nickel(II) nitrate (175 mL, 0.500 M). After the reaction is complete, what is the concentration of the unprecipitated ion (either PO or Ni2+). Na PO, + Ni(NO3)2 — Ni,(PO3)2+ + NaNO,

Answer 1

Ans 1 :

The balanced reaction is given as :

2Na₃PO₄ + 3Ni(NO₃)₂ = Ni₃(PO₄)₂ + 6NaNO₃

Mol sodium phosphate = mass / molar mass

= 12.05 g / 163.94 g/mol = 0.0735 mol

Mol nickel (II) nitrate = molarity x volume (L)

= 0.500 M x 0.175 L

= 0.0875 mol

2 mol sodium phosphate needs 3 mol nickel (II) nitrate

so 0.0735 mol sodium phosphate will need : (0.0735 mol x 3 mol) /2 mol = 0.110 mol nickel (II) nitrate

but here only 0.0875 mol of nickel (II) nitrate is present , so it is the limiting reactant

so number of mol of nickel (II) phosphate formed will be : 0.0875 mol / 3 = 0.029 mol

mass of nickel (II) phosphate = mol x molar mass

= 0.029 mol x 366 g/mol

= 10.68 grams

Mol Na3PO4 utilised = (0.0875 mol x 2 mol) / 3 mol = 0.0583 mol

mol Na3PO4 left excess = 0.0735 mol - 0.0583 mol = 0.0152 mol

so mol of PO₄^3- unprecipitated = 0.0152 mol

concentration of PO₄^3- = mol / volume (L)

= 0.0152 mol / 0.175 L

= 0.0868 M