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2 Pb(s) + 0,(g) 2 PbO(s) An excess of oxygen reacts with 451.4 g of lead, forming 302.8 g of lead(II) oxide. Calculate the percent yield of the reaction.

2 Pb(s) + 0,(g) 2 PbO(s) An excess of oxygen reacts with 451.4 g of lead, forming 302.8 g of lead(II) oxide. Calculate the percent yield of the reaction.

Chemistry
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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**Reaction and Problem Statement** Consider the reaction: \[ 2 \text{Pb (s)} + \text{O}_2 \text{(g)} \rightarrow 2 \text{PbO (s)} \] An excess of oxygen reacts with 451.4 g of lead, forming 302.8 g of lead(II) oxide. Calculate the percent yield of the reaction. --- **Calculation of Percent Yield** To determine the percent yield, use the formula: \[ \text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\% \] 1. **Determine the Theoretical Yield:** Calculate the moles of lead (Pb) using its molar mass (207.2 g/mol): \[ \text{Moles of Pb} = \frac{451.4 \, \text{g}}{207.2 \, \text{g/mol}} \] Since the reaction is \(2 \text{Pb} \rightarrow 2 \text{PbO}\), the moles of lead(II) oxide (PbO) should be the same. Calculate the theoretical yield based on the molar mass of PbO (223.2 g/mol): \[ \text{Theoretical Yield of PbO} = \text{Moles of PbO} \times 223.2 \, \text{g/mol} \] 2. **Calculate the Percent Yield:** Insert the actual yield (302.8 g) and the theoretical yield into the formula above to find the percent yield. --- **Final Answer** Input the calculated percent yield in the provided box: \[\text{percent yield:} \quad \boxed{\phantom{0}} \quad \% \]
Transcribed Image Text:**Reaction and Problem Statement** Consider the reaction: \[ 2 \text{Pb (s)} + \text{O}_2 \text{(g)} \rightarrow 2 \text{PbO (s)} \] An excess of oxygen reacts with 451.4 g of lead, forming 302.8 g of lead(II) oxide. Calculate the percent yield of the reaction. --- **Calculation of Percent Yield** To determine the percent yield, use the formula: \[ \text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\% \] 1. **Determine the Theoretical Yield:** Calculate the moles of lead (Pb) using its molar mass (207.2 g/mol): \[ \text{Moles of Pb} = \frac{451.4 \, \text{g}}{207.2 \, \text{g/mol}} \] Since the reaction is \(2 \text{Pb} \rightarrow 2 \text{PbO}\), the moles of lead(II) oxide (PbO) should be the same. Calculate the theoretical yield based on the molar mass of PbO (223.2 g/mol): \[ \text{Theoretical Yield of PbO} = \text{Moles of PbO} \times 223.2 \, \text{g/mol} \] 2. **Calculate the Percent Yield:** Insert the actual yield (302.8 g) and the theoretical yield into the formula above to find the percent yield. --- **Final Answer** Input the calculated percent yield in the provided box: \[\text{percent yield:} \quad \boxed{\phantom{0}} \quad \% \]
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