A 0.897 g sample containing the chloride ion is treated with excess lead(II) nitrate. The precipitate is filtered using a 0.923 g piece of filter paper. After washing and drying, the mass of the precipitate and filter paper was 2.686 g. c. The number of moles of chloride in the precipitate.
A 0.897 g sample containing the chloride ion is treated with excess lead(II) nitrate. The precipitate is filtered using a 0.923 g piece of filter paper. After washing and drying, the mass of the precipitate and filter paper was 2.686 g. c. The number of moles of chloride in the precipitate.
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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
Transcribed Image Text:**Title: Determining the Moles of Chloride in a Precipitation Reaction**
**Experiment Overview:**
In this experiment, a 0.897 g sample containing chloride ions is treated with an excess of lead(II) nitrate to form a precipitate. This precipitate is filtered using a piece of filter paper weighing 0.923 g. After thorough washing and drying, the combined mass of the precipitate and the filter paper is found to be 2.686 g.
**Objective:**
To calculate the number of moles of chloride ions present in the precipitate.
**Data Analysis:**
1. **Initial Masses:**
- Sample with chloride ions: 0.897 g
- Filter paper: 0.923 g
2. **Final Mass:**
- Precipitate + Filter paper: 2.686 g
3. **Calculating Mass of the Precipitate:**
- Mass of precipitate = Final mass - Mass of filter paper
- = 2.686 g - 0.923 g
- = 1.763 g
4. **Calculating Moles of Chloride:**
- The precipitate formed is lead(II) chloride (PbCl₂).
- Molar mass of PbCl₂ = (207.2 + 2×35.45) g/mol = 278.1 g/mol
- Moles of PbCl₂ = Mass of precipitate / Molar mass of PbCl₂
- = 1.763 g / 278.1 g/mol
- = 0.00634 moles
5. **Number of Moles of Chloride:**
- Each mole of PbCl₂ contains 2 moles of chloride ions.
- Moles of Cl⁻ = 2 × Moles of PbCl₂
- = 2 × 0.00634
- = 0.01268 moles
**Conclusion:**
The experiment determines that the number of moles of chloride in the precipitate is approximately 0.01268 moles. This demonstrates the stoichiometry involved in the precipitation reaction and aids in understanding the quantitative aspects of chemical reactions.
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