You may want to reference (pages 307 - 308) Section 9.3 while completing this pr

Question

You may want to reference (pages 307 - 308) Section 9.3 while completing this problem. A 60.0 mL sample of a 0.112 M potassium sulfate solution is mixed with 39.5 mL of a 0.130 M lead(II) acetate solution and the following precipitation reaction occurs: K_2SO_4(aq)+Pb(C_2H_3O_2)2(aq) rightarrow 2KC_2H_3O_2(aq) + PbSO_4(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. Identify the limiting reactant. 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. Determine the theoretical yield.

Explanation / Answer

nO OF mole of K2SO4 = 0.06*0.112 = 0.00672 mole.

No of mole of Pb(CH3COO)2   = 0.0395*0.13 = 0.00515 mole.

limiting reagent is Pb(CH3COO)2.

No of mole of PbSO4 FORMED = 0.00515*303.2626 = 1.562 grams

practical yield = 1.01 grams

percent yield = practical yield/theoretical yield *100 =

= 1.01/1.562*100 = 64.66%

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