A solution containing 5.00 g of Lead II Nitrate is reacted with 2.50 g of Potass

Question

A solution containing 5.00 g of Lead II Nitrate is reacted with 2.50 g of Potassium Chloride yielding Lead II Chloride, a bright yellow precipitate, and Potassium Nitrate in solution. The precipitate was collected, dried and weighed finding 4.34 g of product. Find the limiting reactant Determine the theoretical yield of PbCl2 Determine the amount of excess r4actant Determine % yield.

Explanation / Answer

2KCl(aq) + Pb(NO3)2(aq) ===> PbCl2(s) + 2KNO3(aq) a) moles of Kcl = 2.5/74.5=0.03355 moles of Pb(NO3)2 = 5/331.2 =0.01509 for 1 mole lead nitrate 2 moles kcl are required for 0.01509 moles PbNO32moles of KCL reuired= 2*0.01509=0.03018 we have excess of Kcl as we have 0.03355 moles of KCl so LIMNITING REAGENT=Pb(NO3)2 B. thoretical yioeld= 0.01509 moles C.amount of excess reactnat= 0.03355-0.03018=0.00337 moles =0.251 grams of KCl D percentage yield = 0.01509/0.03355 *100 =44.9 %

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