Question

You have 315 mL of an aqueous solution that is 0.29 M Na2SO4 and a separate 250 mL aqueous solution that is 0.41 M BaCl2- a). How many moles of each of the four ions are present (do not use the word "moles" in your answer, just the number)? Na+ 5042- Ba2+ ci b). When these two solutions are mixed together, what is the chemical formula of the solid expected to precipitate from the mixture? (Don't forget to put (s) into ChemPad immediately after the compound formula without any spaces between) Help chemPad X.Xº =+ Greek c). If the sulfate ions in the mixed solution react with the barium ions in the mixed solution to form a precipitate, how many moles of Ba2+ ions are required to react with all of the SO42- ions present? Express your answer in scientific notation using the format 1.23e+45 or 1.23e-45. moles of Ba2+

Answer 1

a)

Molarity=number of moles/Volume(in L)

number of moles= Molarity=Volume(in L)

For Na₂SO₄ , Volume=315 mL=0.315L & Molarity=0.29 M

number of moles of Na₂SO₄ =0.29 x 0.315 moles=0.09135 moles

For BaCl₂ , Volume=250 mL=0.25 L & Molarity=0.41 M

number of moles of Na₂SO₄ =0.25 x 0.41 moles=0.1025 moles

Assuming complete dissociation:

One molecule of Na₂SO₄ gives two Na⁺ ions and two SO₄^2- ions &

One mole of BaCl₂ gives one mole of Ba²⁺ ions and two moles of Cl^- ions

Therefore

Number of moles of Na⁺ ions=2 x 0.09135 moles=0.1827 moles

Number of moles of SO₄^2- ions=0.09135 moles

                                             &

Number of moles of Ba⁺² ions=0.1025 moles

Number of moles of Cl^- ions=0.205 moles

b)

BaCl(aq) + Na2SO4(aq) — BaSO4(s) +2 NaCl(aq)

BaSO₄(s) is the precipitate.

c) To completely precipitate out SO₄^2- ions, equal number of mole of Ba²⁺ ions are required

     which is equal to 0.09135 moles=9.135e-2 moles.

d) To completely precipitate out Ba²⁺ ions, equal number of mole of SO₄^2- ions are required

     which is equal to 0.09135 moles=0.1025 moles=1.025e+1 moles.

e) Number moles of sulfate ions are lower than that of the Barium ions, hence sufate ion is the limiting reagent.

f) Since sufate ion is the limiting reagent, number of moles of precipitate, BaSO₄(s) formed=number of moles of sulfate ion=0.09135 moles

number of moles=amount(in grams)/molar mass(in grams/Mole)

Molar mass of BaSO₄(s)= 233.38 g/mol

Amount of BaSO4(s)= 233.38 g/mol x 0.09135 moles= 21.32 g

g) Spectator cation- Na⁺

   Spectator anion- Cl^-

h) NaCl is formed along with precipitate, which is a strong electrolyte.