Question

Nickel (II) chloride reacts with sodium phosphate to form solid nickel (II) phosphate and aqueous sodium chloride. (a) Determine the percent yield if a student isolated 67.7 mg of solid nickel (II) phosphate from the following reaction mixture: 0.207 g nickel (II) chloride and 0.186 g sodium phosphate. (b) What mass, in grams, of excess reagent remains?

Answer 1

a)
3 NiCl2 (aq) + 2 Na3PO4 (aq) = Ni3(PO4)2 (s) + 6 NaCl (aq)

number of moles of NiCl2 reacted = mass/molar mass
= 0.207/129.6
= 0.0016 mol

number of moles of Na3PO4 reacted = mass/molar mass
= 0.186/164
= 0.0011 mol

0.0011 mol of Na3PO4 will need (3/2)*0.0011 = 0.0017 mol of NiCl2 to react completely
so,
NiCl2 is limiting reagent

theoretically, moles of Ni3(PO4)2 formed = 0.0016/3 = 0.000533 mol
mass of Ni3(PO4)2 formed = number of mols * molar mass
= 0.000533 * 366
= 0.1952 g
=195.2 mg

in actual mass produced = 67.7 mg

percent yield = 67.7*100/195.2
= 34.7 %
Answer: 34.7 %

b)
number of moles of Na3PO4 remaining = 0.0011 - (2/3)*0.0016 mol = 3.33*10^-5 mol
mass of Na3PO4 remaining = 3.33*10^-5 * 164 =5.47*10^-3 g = 5.47 mg
Answer: 5.47 mg