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11Alexandr11
1 month ago
12

A 25.0-mL sample of a 1.20 M potassium chloride solution is mixed with 15.0 mL of a 0.900 M lead(II) nitrate solution and this p

recipitation reaction occurs: 2 KCl(aq)+Pb(NO3)2(aq)→PbCl2(s)+2 KNO3(aq) The solid PbCl2 is collected, dried, and found to have a mass of 2.45 g. Determine the limiting reactant, the theoretical yield, and the percent yield.
Chemistry
1 answer:
Alekssandra [3K]1 month ago
6 0
The limiting reagent is lead(II) nitrate, with a theoretical yield of 3.75435 g, resulting in a percent yield of 65.26%. The calculated moles involved in the reaction reflect that 1.20 M potassium chloride is present in 25.0 mL, which converts to 0.03 moles. In contrast, 0.900 M lead(II) nitrate in 15.0 mL results in 0.0135 moles. The limiting reagent, being the one present in smaller quantities, drives the reaction yielding 0.0135 moles of lead(II) chloride, correlating to a mass of 3.75435 g. Finally, given an experimental yield of 2.45 g, the percent yield calculation leads to 65.26%.
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