Question

You mix 6.0mL of 0.50M Na₂SO₄ with 8.0mL of 0.50M Pb(NO₃)₂ and collect and dry the PbSO₄ precipitate. Calculate the following:

a. Moles of Na₂SO₄ used

b. Moles of Pb(NO₃)₂ used

c. Moles of PbSO₄ precipitate expected from Na₂SO₄

d. Moles of PbSO₄ precipitate expected from Pb(NO₃)₂

e. What mass of PbSO₄ do you expect to obtain from this reaction?

f. If the actual mass of the precipitate you recover is 0.78g, what is the percent recovery of the precipitate?

Answer 1

a)

moles of NA2SO4 can be calculated via molarity and volume:

Mol of Na2SO4 = M*V = 6*0.5 = 3 mmol of Na2SO4

3*10^-3 mol of Na2SO4

b)

moles of Pb(NO3)2 can be calculated via molarity and volume:

Mol of Pb(NO3)2 = MV = 8*0.5 = 4 mmol of Pb(NO3)2

4*10^-3 mol of Pb(NO3)2

c)

you will expect from Na2SO4:

1:1 ratio (1 mol of SO4-2 in PbSO4 and 1 mol of SO4-2 in NA2SO3)

3*10^-3 mol of Na2SO4 = 3*10^-3 mol of PbSO4

d)

you will expect from Pb(NO3)2:

1:1 ratio (1 mol of Pb in PbSO4 and 1 mol of PB in Pb(NO3)2)

4*10^-3 mol of Pb(NO3)2 = 4*10^-3 mol of PbSO4

e)

what mass of PbSO4 will be obtained

there is limiting reactnat, which is NA2SO4, i.e. SO4-2 limits

so

3*10^-3 mol of SO4-2 form #*10^-3 mol of PBSO4

mass = mol*M W= (3*10^-3)(303.26 ) = 0.90978 g of PbSO4 g are expected

f)

yield = real / theoretical * 100% = 0.78/0.90978*100 = 85.735 %