A 61.0 mL sample of a 0.104 M potassium sulfate solution is mixed with 36.0 mL o

Question

A 61.0 mL sample of a 0.104 M potassium sulfate solution is mixed with 36.0 mL of a 0.106 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 0.992 g .

Determine the limiting reactant, the theoretical yield, and the percent yield.

Part A

Identify the limiting reactant.

Identify the limiting reactant.

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The limiting reactant of a reaction is found by calculating the number of moles of each reactant and comparing them. The reactant with the least number of moles is considered the limiting reactant because once all of it has reacted the reaction stops despite the presence of another reactant. The limiting reactant is important when determining the theoretical yield of a product.

The limiting reactant is solely applied to the reactants in the equation, and not the products. Since PbSO4 is a product so it cannot be the limiting reactant.

Part B

Determine the theoretical yield.

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Part C

Determine the percent yield.

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Explanation / Answer

A)

Moles of K2SO4 = M*V
= 0.104 M * 0.061 L
= 0.006344 mol

moles of Pb(C2H3O2)2 = M*V
= 0.106 M * 0.036 L
=0.003816 mol
Pb(C2H3O2)2 is the limiting reagent

B)
moles of PbSO4 formed = 0.003816 mol
molar mass of PbSO4 = 303.26 g/mol
mass of PbSO4 = number of moles * molar mass
= 0.003816 * 303.26
= 1.157 g
Answer: 1.157 g

C)
actual yield = 0.992 g
percent yield = 0.992*100 / 1.157
= 85.7 %
Answer: 85.7 %

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