Question

If 2.000 g of Na3 PO4 •12H 2 O is reacted with 2.000 g of SrCl2 ·6H 2 O

Calculate the theoretical yield of Sr3 (PO4 )2 produced from 2.000 g of Na3 PO 4 •12H 2 O if we assume that the strontium chloride salt is in excess.

Calculate the theoretical yield of Sr3 (PO4 )2 produced from 2.000 g of SrCl2 ·6H 2 O again assuming that the sodium phosphate is in excess.

Which reactant produces the smallest theoretical yield?

Which compound above is the limiting reactant?

Answer 1

mol of SrCl2 ·6H 2 O = mass/MW = 2/(158.526 + 6*18) = 0.0075

mol of Na3 PO4 •12H 2 O = mass/MW = 2/(163.940671 + 12*18) = 0.005263

a)

if SrCl2 salt is in ecess

1 mol of SrCl2 --> 1/3 mol of Sr3(PO4)2 --> 1/3*0.0075 = 0.0025 mol of salt will produce Sr salt

b)

if PO4-3 salt is in ecess

1 mol of Na3PO4 --> 1/2 mol of Sr3(PO4)2 --> 1/2*0.005263 = 0.0026315 mol of salt will produce Sr salt

c)

smallest yield -->  0.0025 mol of salt will produce Sr salt

d)

this is the limiting reactants, that is,  2.000 g of SrCl2 ·6H 2 O is limiting, will react all first