Question

What are the possible theoretical yields (from both reactants) of NaCl if a solution containing 33.40 grams of sodium phosphate produced 19.60 g of AlPO4(s) when reacted with 33.40 g aluminum chloride in solution? The chemical formula MAY be unbalanced!

(Na = 22.99 amu, P = 30.97 amu, O = 16.00 amu, Al = 26.98 amu, Cl = 35.45 amu)

Na3PO4(aq) + AlCl3(aq) → NaCl(aq) + AlPO4(s)

Theoretical yield of NaCl from Na3PO4 = Blank 1 grams of NaCl

Theoretical yield of NaCl from AlCl3 = Blank 2 grams of NaCl

Answer 1

The balanced equation is

Na3PO4 (aq) + AlCl3 (aq) ---------------> 3 NaCl (aq) + AlPO4 (s)

a) from 33.40g of Na3PO4

1 mol of Na3PO4 gives 3 moles of NaCl

94.97g/mol of Na3PO4 gives 3x 58.44g/mol of NaCl

33.40g of Na3PO4 can give = [33.40g x 3 x 58.44g/mol ] / 94.97 g/mol

= 61.65 g of NaCl

Thus the theoretical yield of NaCl from Na3PO4 = 61.65 g

b) From stoichiometric equation

1 moles of AlCl3 produces 3 moles of NaCl

133.33 g/mol of AlCl3 gives 3 x 58.44 g/mol of NaCl

33.40 g of AlCL3 gives = 33.40g 3x 58.44g/mol / 133.33 g/mol

= 43.92 g of NaCl

Thus theoretical yield from ALcL3 = 43.92 g of NaCl