A solution of calcium chloride (100.0 mL, 1.00 M) is mixed with ammonium phosphate (100.0 mL, 1.00 M). Calculate the concentration of all ions in solution (assume 200.0 mL total volume) and determine the mass of precipitate formed in the reaction.
A solution of calcium chloride (100.0 mL, 1.00 M) is mixed with ammonium phosphate (100.0 mL, 1.00 M). Calculate the concentration of all ions in solution (assume 200.0 mL total volume) and determine the mass of precipitate formed in the reaction.
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
Chapter1: Chemical Foundations
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![### Chemistry Problem: Mixture of Calcium Chloride and Ammonium Phosphate
**Problem Statement:**
A solution of calcium chloride (100.0 mL, 1.00 M) is mixed with ammonium phosphate (100.0 mL, 1.00 M). Calculate the concentration of all ions in solution (assume 200.0 mL total volume) and determine the mass of precipitate formed in the reaction.
**Detailed Steps and Solutions:**
1. **Write the balanced chemical equation for the reaction:**
\[ \text{3CaCl}_2 + \text{2(NH}_4\text{)_3PO}_4 \rightarrow \text{Ca}_3(\text{PO}_4)_2 (\text{s}) + 6 \text{NH}_4\text{Cl} \]
2. **Calculate the moles of each reactant:**
- For Calcium Chloride (CaCl₂): \[ \text{Moles of CaCl}_2 = (1.00 \, \text{M}) \times (0.100 \, \text{L}) = 0.100 \, \text{moles} \]
- For Ammonium Phosphate [(NH₄)₃PO₄]: \[ \text{Moles of (NH}_4\text{)_3PO}_4 = (1.00 \, \text{M}) \times (0.100 \, \text{L}) = 0.100 \, \text{moles} \]
3. **Determine the limiting reagent:**
By using stoichiometry, we need 3 moles of CaCl₂ for every 2 moles of (NH₄)₃PO₄.
- Moles of CaCl₂ required for 0.100 moles of (NH₄)₃PO₄:
\[ \text{Required moles of CaCl}_2 = 0.100 \, \text{moles} \times \left( \frac{3 \text{ moles CaCl}_2}{2 \text{ moles } (\text{NH}_4)_3\text{PO}_4} \right) = 0.150 \text{ moles CaCl}_2 \]
Since we only have 0.100 moles](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd840dc46-e3d2-47d1-9b6d-e93b2d8faa77%2F82bdbab1-d2c6-4de8-bbea-e12ecef6693d%2Fa5m2pp7.jpeg&w=3840&q=75)
Transcribed Image Text:### Chemistry Problem: Mixture of Calcium Chloride and Ammonium Phosphate
**Problem Statement:**
A solution of calcium chloride (100.0 mL, 1.00 M) is mixed with ammonium phosphate (100.0 mL, 1.00 M). Calculate the concentration of all ions in solution (assume 200.0 mL total volume) and determine the mass of precipitate formed in the reaction.
**Detailed Steps and Solutions:**
1. **Write the balanced chemical equation for the reaction:**
\[ \text{3CaCl}_2 + \text{2(NH}_4\text{)_3PO}_4 \rightarrow \text{Ca}_3(\text{PO}_4)_2 (\text{s}) + 6 \text{NH}_4\text{Cl} \]
2. **Calculate the moles of each reactant:**
- For Calcium Chloride (CaCl₂): \[ \text{Moles of CaCl}_2 = (1.00 \, \text{M}) \times (0.100 \, \text{L}) = 0.100 \, \text{moles} \]
- For Ammonium Phosphate [(NH₄)₃PO₄]: \[ \text{Moles of (NH}_4\text{)_3PO}_4 = (1.00 \, \text{M}) \times (0.100 \, \text{L}) = 0.100 \, \text{moles} \]
3. **Determine the limiting reagent:**
By using stoichiometry, we need 3 moles of CaCl₂ for every 2 moles of (NH₄)₃PO₄.
- Moles of CaCl₂ required for 0.100 moles of (NH₄)₃PO₄:
\[ \text{Required moles of CaCl}_2 = 0.100 \, \text{moles} \times \left( \frac{3 \text{ moles CaCl}_2}{2 \text{ moles } (\text{NH}_4)_3\text{PO}_4} \right) = 0.150 \text{ moles CaCl}_2 \]
Since we only have 0.100 moles
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