Chemistry
Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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### Precipitation of Lead Ions

Lead 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 Details

When 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 g

2. **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:**

\[
\text{\% yield} = \boxed{ \quad } \%
\]
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Transcribed Image Text:### Precipitation of Lead Ions Lead 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 Details When 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 g 2. **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 heating
magnesium metal in the presence of the oxygen. The
balanced 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 MgO
is collected.
The theoretical yield is determined from the limiting reactant. The mass of Mg
moles of MgO to grams of MgO.
- Part C
Determine the percent yield for the reaction.
Express your answer as a percent.
[ΨΕΙ ΑΣΦ
Ú
?
%
2
expand button
Transcribed Image Text:Magnesium oxide can be made by heating magnesium metal in the presence of the oxygen. The balanced 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 MgO is collected. The theoretical yield is determined from the limiting reactant. The mass of Mg moles of MgO to grams of MgO. - Part C Determine the percent yield for the reaction. Express your answer as a percent. [ΨΕΙ ΑΣΦ Ú ? % 2
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