Question

A 55.0 mL sample of a 0.102 M potassium sulfate solution is mixed with 35.0 mL of a 0.114 M lead (II) acetate solution. The solid lead (II) sulfate is collected, dried, and found to have a mass of 1.01 g.

The Balanced Reaction is: Pb(CH3COO2)(aq) + K2SO4 --> PbSO4(S) + 2K (aq) + 2CH3COO- (aq)

b.Write the net ionic reaction.

c.Which ions are spectators?

d.Determine the limiting reagent.

e. Determine the theoretical yield.

f.Determine the percent yield.

g. Determine the concentration of the remaining ios in solution (spectator AND excess reagent ions.)

Answer 1

b. Pb2+ (aq) + SO4 2-(aq) ---> PbSO4(S)

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Pb2+ (aq) + 2 CH3COO-(aq) + 2 K+ (aq) + SO4 2- (aq) ---> PbSO4(S) + 2 K+ (aq) + 2 CH3COO- (aq)

removing spectator ions, net ionic reaction

c. Spectator ions, CH3COO-, K+

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Spectator ions does not take part in the reaction, they appear on both sides of the equation

d. Limiting reagent, Pb(CH3COO)2

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No. of moles of 0.102 M, 55.0 mL K2SO4 = 0.102 x 55.0 mL/1000 = 0.00561 mol

No. of moles of 0.114 M, 35.0 mL Pb(CH3COO)2 = 0.114 M x 35.0 mL/1000 = 0.00399 mol

Since the reaction is 1:1, the rate limiting reagent will be the one which has least concentration in the reaction mixture, i.e. Pb(CH3COO)2 (0.00399 mol)

e. Theoretical yield, 1.210 g

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Will be equal the about of least available reagent in the reaction, i.e. 0.00399 mol.

M.W. of PbSO4 = 303.26 g/mol

So yield = 0.00399 mol x 303.26 g/mol = 1.210 g

f. Percentage yield, 83.47 %

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Found mass of PbSO4 = 1.01 g

Theoretical mass = 1.210 g

Yield (%) = 1.01 g x 100/1.210 = 83.47 %

g. Concentration of remaining reagents,

K2SO4 = 0.00228 mol = 0.3973 g = 0.0002052 M

Pb(CH3COO)2 = 0.00066 mol = 0.2147 g = 0.0000594 M

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Since the reaction is 1:1, 0.00399 mol of Pb(CH3COO)2 reacts 0.00399 mol of K2SO4, so the amount of K2SO4 in excess = 0.00561 mol - 0.00399 mol = 0.00162 mol.

Excess K2SO4 = 0.00162 mol.

Since the reaction is only 83.47 %, 16.53 % of 0.00399 mol of both Pb(CH3COO)2 and K2SO4 are left unreacted

i.e. 0.00399 x 16.53/100 = 0.00066 mol

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Unreacted K2SO4 = 0.00066 mol

Total unreacted K2SO4 = 0.00162 mol + 0.00066 mol = 0.00228 mol

M.W. K2SO4 = 174.259 g/mol

Mass of unreacted K2SO4 = 0.00228 mol x 174.259 g/mol = 0.3973 g

Unreacted Pb(CH3COO)2 = 0.00066 mol

M.W. of Ph(CH3COO)2 = 325.29 g/mol

Mass of unreacted Ph(CH3COO)2 = 0.00066 mol x 325.29 g/mol = 0.2147 g

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Total volume = 55.0 mL + 35.0 mL = 90 mL

Molarity of unreacted K2SO4 in 90 mL solution = 0.00228 mol x 90/1000 = 0.0002052 M

Molarity of unreacted Ph(CH3COO)2 in 90 mL solution = 0.00066 mol x 90/1000 = 0.0000594 M

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