Question

1. Balance the reaction below and determine how many grams of nickel(II) phosphate will be formed when sodium phosphate (12.0
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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 PO43- unprecipitated = 0.0152 mol

concentration of PO43- = mol / volume (L)

= 0.0152 mol / 0.175 L

= 0.0868 M

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