### Precipitation of Lead IonsLead ions can be precipitated from a solution with KCl according to the reaction:\[ \text{Pb}^{2+} \text{(aq)} + 2\text{KCl(aq)} \rightarrow \text{PbCl}_2 \text{(s)} + 2\text{K}^+ \text{(aq)} \]### Reaction DetailsWhen 34.3 g of KCl is added to a solution containing 25.8 g of \(\text{Pb}^{2+}\), PbCl₂(s) forms. The solid is filtered and dried, and it is found to have a mass of 30.9 g.### Calculating Percent Yield**Part C: Determine the percent yield for the reaction.**Express your answer in percent to three significant figures.#### Formula for Percent Yield\[ \text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100 \]#### Given Data- Actual Yield: 30.9 g- **Theoretical Yield** (provided above): 25.8 g**Calculation:**1. **Identify the masses of reagents and products:** - Actual yield of PbCl₂(s): 30.9 g - Theoretical yield of PbCl₂(s): 25.8 g2. **Determine the percent yield:** \[ \text{Percent Yield} = \left( \frac{30.9 \, \text{g}}{25.8 \, \text{g}} \right) \times 100 \approx 119.77\% \]3. **Round to three significant figures:** \[ \text{Percent Yield} \approx 119.8\% \]**Inputs for Calculation:**\[\text{\% yield} = \boxed{ \quad } \%\] Magnesium oxide can be made by heatingmagnesium metal in the presence of the oxygen. Thebalanced equation for the reaction is:2Mg(s) + O2(g) → 2MgO(s)When 12.9 g Mg reacts with 14.0 g O2, 12.5 g MgOis collected.The theoretical yield is determined from the limiting reactant. The mass of Mgmoles of MgO to grams of MgO.- Part CDetermine the percent yield for the reaction.Express your answer as a percent.[ΨΕΙ ΑΣΦÚ?%2

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Lead ions can be precipitated from solution with KCl according to the reaction: Pb²+ (aq) + 2KCl(aq) → PbCl₂ (s) + 2K+ (aq) When 34.3 g KCl is added to a solution containing 25.8 g Pb2+, PbCl2 (s) forms. The solid is filtered and dried and found to have a mass of 30.9 g. theoretical yield of PbCl2 = 25.8 g Ph Part C % yield = X Determine the percent yield for the reaction. Express your answer in percent to three significant figures. ΠΫΠΙ ΑΣΦ 1 mol Pb2+ 1 mol 1 met X 207.2 g Pb²+ 2 ? %

Lead ions can be precipitated from solution with KCl according to the reaction: Pb²+ (aq) + 2KCl(aq) → PbCl₂ (s) + 2K+ (aq) When 34.3 g KCl is added to a solution containing 25.8 g Pb2+, PbCl2 (s) forms. The solid is filtered and dried and found to have a mass of 30.9 g. theoretical yield of PbCl2 = 25.8 g Ph Part C % yield = X Determine the percent yield for the reaction. Express your answer in percent to three significant figures. ΠΫΠΙ ΑΣΦ 1 mol Pb2+ 1 mol 1 met X 207.2 g Pb²+ 2 ? %

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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