A 55.0 mL sample of a 0.104 M potassium sulfate solution is mixed with 37.5 mL o

Question

A 55.0 mL sample of a 0.104 M potassium sulfate solution is mixed with 37.5 mL of a 0.106 M lead(II) acetate solution and the following precipitation reaction occurs: K_2SO_4(aq) + Pb(C_2H_3O_2)_2(aq)rightarrow2KC_2H_3O_2(aq) + PbSO_4(s) The solid PbSO_4 is collected, dried, and found to have a mass of 0.998 g. Determine the limiting reactant, the theoretical yield, and the percent yield. Identify the limiting reactant. Pb(C_2H_3O_2)_2 PbSO_4 K_2SO_4 KC_2H_3O_2 Determine the theoretical yield. Determine the percent yield.

Explanation / Answer

K2SO4 moles = M x V ( inL)

                   = 0.104 x ( 0.055) = 0.00572

Pb(C2H3O2)2 moles = 0.106 x 0.0375 = 0.003975

both reactaats react n 1:1 ratio but since Pb(C2H3O2)2 is relatively less it is limiting reagent

Limiting reagent is Pb(C2H3O2)2

Theoretical yield of PbSO4 moles = Pb(C2H3O2)2 moles reacted = 0.003975

PbSO4 mass = moles x molar mass of PbSO4

         = 0.003975 mol x 303.26 g/mol = 1.2 g

thus theoretical yiled PbSO4 mass= 1.2 g

actual yiled = 0.998 g

thus percent yiled = ( 100 x actual mass/ theoretical yield)

        = ( 100 x 0.998 /1.2) = 83.2 %

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