A 62.0 mL sample of a 0.112 M potassium sulfate solution is mixed with 34.0 mL of a 0.118 M lead(II) acetate solution and the following precipitation reaction occurs: K2SO4(aq)+Pb(C2H3O2)2(aq)→2KC2H3O2(aq)+PbSO4(s) The solid PbSO4 is collected, dried, and found to have a mass of 1.00 g . Determine the limiting reactant, the theoretical yield, and the percent yield.

To determine the limiting reactant, we need to compare the number of moles of each reactant.

First, let's calculate the number of moles of potassium sulfate (K2SO4) in the 62.0 mL solution. We can use the formula:

moles = concentration (M) x volume (L)

Concentration of K2SO4 = 0.112 M Volume of K2SO4 solution = 62.0 mL = 0.0620 L

moles of K2SO4 = 0.112 M x 0.0620 L = 0.00694 mol

Next, let's calculate the number of moles of lead(II) acetate (Pb(C2H3O2)2) in the 34.0 mL solution:

Concentration of Pb(C2H3O2)2 = 0.118 M Volume of Pb(C2H3O2)2 solution = 34.0 mL = 0.0340 L

moles of Pb(C2H3O2)2 = 0.118 M x 0.0340 L = 0.00401 mol

According to the balanced equation, the stoichiometric ratio between K2SO4 and Pb(C2H3O2)2 is 1:1. This means that for every 1 mole of K2SO4, we need 1 mole of Pb(C2H3O2)2.

Since the number of moles of Pb(C2H3O2)2 (0.00401 mol) is less than the number of moles of K2SO4 (0.00694 mol), Pb(C2H3O2)2 is the limiting reactant.

To determine the theoretical yield of PbSO4, we need to calculate the number of moles of PbSO4 that can be formed from the limiting reactant. According to the balanced equation, the stoichiometric ratio between Pb(C2H3O2)2 and PbSO4 is 1:1.

Therefore, the number of moles of PbSO4 formed is also 0.00401 mol.

To convert the moles of PbSO4 to grams, we can use the molar mass of PbSO4, which is 303.26 g/mol.

mass of PbSO4 = moles of PbSO4 x molar mass of PbSO4 mass of PbSO4 = 0.00401 mol x 303.26 g/mol = 1.22 g

The theoretical yield of PbSO4 is 1.22 g.

To calculate the percent yield, we need to compare the actual yield (1.00 g) to the theoretical yield (1.22 g).

percent yield = (actual yield / theoretical yield) x 100 percent yield = (1.00 g / 1.22 g) x 100 = 81.97%

Therefore, the limiting reactant is Pb(C2H3O2)2, the theoretical yield of PbSO4 is 1.22 g, and the percent yield is 81.97%.