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 and this precipitation reaction occurs:

K2SO4(aq) + Pb(C2H3O2)2(aq) → 2 KC2H3O2(aq) + PbSO4(s)

The solid PbSO4 is collected, dried, and found to have a mass of 1.01 g. Determine the limiting reactant, the theoretical yield, and the percent yield.

Answer 1

To determine the limiting reactant, compare the number of moles of each reactant to the stoichiometric ratio. Pb(C2H3O2)2 is the limiting reactant. The theoretical yield of PbSO4 is calculated using the mole ratio between Pb(C2H3O2)2 and PbSO4. The per cent yield is calculated using the actual yield and theoretical yield.

Step-by-step explanation:

To determine the limiting reactant, we need to compare the amount of each reactant to the stoichiometric ratio in the balanced equation. From the given information, the number of moles of potassium sulfate (K2SO4) can be calculated:

Moles of K2SO4 = (volume of solution in liters) x (molarity of K2SO4)

Moles of K2SO4 = 0.055 L x 0.102 M = 0.00561 moles

Similarly, the number of moles of lead(II) acetate (Pb(C2H3O2)2) can be calculated:

Moles of Pb(C2H3O2)2 = (volume of solution in liters) x (molarity of Pb(C2H3O2)2)

Moles of Pb(C2H3O2)2 = 0.035 L x 0.114 M = 0.00399 moles

Since the stoichiometric ratio in the balanced equation is 1:1, the reactant with the smaller number of moles is the limiting reactant. In this case, Pb(C2H3O2)2 is the limiting reactant.

The theoretical yield of PbSO4 can be calculated using the mole ratio between Pb(C2H3O2)2 and PbSO4:

Moles of PbSO4 = (moles of Pb(C2H3O2)2) x (molar mass of PbSO4 / molar mass of Pb(C2H3O2)2)

Moles of PbSO4 = 0.00399 moles x (239.2 g/mol / 325.3 g/mol) = 0.00293 moles

The molar mass of PbSO4 is 239.2 g/mol and the molar mass of Pb(C2H3O2)2 is 325.3 g/mol.

The actual yield of PbSO4 is given as 1.01 g. The percent yield can be calculated using:

Percent yield = (actual yield / theoretical yield) x 100%

Percent yield = (1.01 g / (0.00293 moles x 239.2 g/mol)) x 100% = 115%

Answer 2

The limiting reactant is lead (II) acetate, the theoretical yield is 1.21 grams of PbSO4, and the percent yield is approximately 83.5%.

Step-by-step explanation:

Determining Limiting Reactant, Theoretical Yield, and Percent Yield

To determine the limiting reactant, first calculate the moles of each reactant. The compound of which the fewer moles are present will be the limiting reactant, as it will be completely consumed first during the reaction. Use the stoichiometry of the balanced equation to relate the moles of the limiting reactant to the moles of the product, lead (II) sulfate (PbSO4).

For potassium sulfate (K2SO4):

Moles = 0.055 L × 0.102 mol/L = 0.00561 mol

For lead (II) acetate (Pb(C2H3O2)2):

Moles = 0.035 L × 0.114 mol/L = 0.00399 mol

From the balanced equation, it takes 1 mole of K2SO4 to react with 1 mole of Pb(C2H3O2)2, so lead (II) acetate is the limiting reactant as it has fewer moles.

The theoretical yield is the mass of PbSO4 that could be formed from the limiting reactant if the reaction were 100% efficient. Since the molar mass of PbSO4 is 303.26 g/mol, the theoretical yield can be calculated as follows:

Theoretical yield = Moles of PbSO4 × Molar mass of PbSO4 = 0.00399 mol × 303.26 g/mol = 1.21 g

The percent yield is the actual yield divided by the theoretical yield, multiplied by 100%:

Percent Yield = (Actual yield / Theoretical yield) × 100% = (1.01 g / 1.21 g) × 100% ≈ 83.5%

Therefore, the limiting reactant is lead (II) acetate, the theoretical yield is 1.21 grams of PbSO4, and the percent yield is approximately 83.5%.