A 61.0 mL sample of a 0.118 M potassium sulfate solution is mixed with 35.0 mL o

Question

A 61.0 mL sample of a 0.118 M potassium sulfate solution is mixed with 35.0 mL of a 0.114 M lead(II) acetate solution and the following precipitation reaction occurs: K2SO4 (aq) 2)2 (aq)- 2KC2H3O2 (aq) Pbso4 (s) The solid PbSO 4 is collected, dried, and found to have a mass of 1.01 g Determine the limiting reactant, the theoretical yield, and the percent yield. Part A Identify the limiting reactant. K2 so 2) 2 O PbSO4 KC Hao Submit My Answers Give Up Part B Determine the theoretical yield. mass of PbSO4 Submit My Answers Give Up Part C Determine the percent yield.

Explanation / Answer

Limiting reagent is defined as the reagent which yields least amount of product in a reaction.

Find moles of each reactant and calculate the theoretical yield of product formed from each. The lowest would be the limiting reagent.

Part A : The limiting reagent

moles of K2SO4 = M x L = 0.118 M x 0.061 L = 7.198 x 10^-3 mols

moles of Pb(OAc)2 = 0.114 M x 0.035 L = 3.99 x 10^-3 mols

molar mass of PbSO4 = 303.26 g/mol

weight of PbSO4 formed from K2SO4 = 7.198 x 10^-3 x 303.26 = 2.18 g

weight of PbSO4 formed from Pb(OAc)2 = 3.99 x 10^-3 x 303.26 = 1.21 g

thus, limiting reagent here would be Pb(OAc)2

Part B : Theoretical yield of PbSO4 = 1.21 g

Part C : Percent yield of PbSO4 = (1.01 / 1.21) x 100 = 83.5 %

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