reaction between lead (II) nitrate and potassium iodide results in 35.0 g (actual yield) of the yellow precipitate. Determine the percent yield of the reaction between 150.0 mL of 0.800 mol/L lead (II) nitrate and 200.0 mL of 1.00 mol/L potassium iodide. ___ Pb(NO3)2 (aq) + __2_ KI (aq) ---> __2_ KNO3 (aq) + ___PbI2(s)
reaction between lead (II) nitrate and potassium iodide results in 35.0 g (actual yield) of the yellow precipitate. Determine the percent yield of the reaction between 150.0 mL of 0.800 mol/L lead (II) nitrate and 200.0 mL of 1.00 mol/L potassium iodide. ___ Pb(NO3)2 (aq) + __2_ KI (aq) ---> __2_ KNO3 (aq) + ___PbI2(s)
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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- A reaction between lead (II) nitrate and potassium iodide results in 35.0 g (actual yield) of the yellow precipitate. Determine the percent yield of the reaction between 150.0 mL of 0.800 mol/L lead (II) nitrate and 200.0 mL of 1.00 mol/L potassium iodide.
___ Pb(NO3)2 (aq) + __2_ KI (aq) ---> __2_ KNO3 (aq) + ___PbI2(s)
- Find the limiting reagent.
- Find the theoretical yield.
- Find the percentage yield.
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