2. A binary mixture of 0.437 g BaCl2•2H2O (molar mass = 244.27 g/mol) and 0.284 g Na2SO4 (molar mass = 142.04 g/mol) is added to water. What is the theoretical yield in grams of BaSO4 precipitate (molar mass = 233.39 g/mol)?
2. A binary mixture of 0.437 g BaCl2•2H2O (molar mass = 244.27 g/mol) and 0.284 g Na2SO4 (molar mass = 142.04 g/mol) is added to water. What is the theoretical yield in grams of BaSO4 precipitate (molar mass = 233.39 g/mol)?
Chemistry
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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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![**Problem 2: Theoretical Yield Calculation**
A binary mixture of 0.437 g BaCl₂•2H₂O (molar mass = 244.27 g/mol) and 0.284 g Na₂SO₄ (molar mass = 142.04 g/mol) is added to water. What is the theoretical yield in grams of BaSO₄ precipitate (molar mass = 233.39 g/mol)?
**Solution Steps:**
1. Calculate the number of moles of BaCl₂•2H₂O:
\[ \text{Moles of BaCl₂•2H₂O} = \frac{\text{Mass of BaCl₂•2H₂O}}{\text{Molar Mass of BaCl₂•2H₂O}} = \frac{0.437\, \text{g}}{244.27\, \text{g/mol}} \]
2. Calculate the number of moles of Na₂SO₄:
\[ \text{Moles of Na₂SO₄} = \frac{\text{Mass of Na₂SO₄}}{\text{Molar Mass of Na₂SO₄}} = \frac{0.284\, \text{g}}{142.04\, \text{g/mol}} \]
3. Determine the limiting reagent by comparing the mole ratio between BaCl₂ and Na₂SO₄.
4. Use the moles of the limiting reagent to calculate the theoretical yield of BaSO₄.
5. The reaction equation:
\[ \text{BaCl₂ + Na₂SO₄ → BaSO₄ + 2 NaCl} \]
6. Calculate the mass of BaSO₄ precipitate:
\[ \text{Mass of BaSO₄} = \text{Moles of BaSO₄} \times \text{Molar Mass of BaSO₄} \]
By following these steps, you will determine the theoretical yield of BaSO₄ precipitate in grams.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0a4a1727-156d-4ade-b12f-025bad03a8ba%2F3b71b3e6-cde6-43bd-9530-7f3457e1c5d5%2Flc5s0zn0e_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem 2: Theoretical Yield Calculation**
A binary mixture of 0.437 g BaCl₂•2H₂O (molar mass = 244.27 g/mol) and 0.284 g Na₂SO₄ (molar mass = 142.04 g/mol) is added to water. What is the theoretical yield in grams of BaSO₄ precipitate (molar mass = 233.39 g/mol)?
**Solution Steps:**
1. Calculate the number of moles of BaCl₂•2H₂O:
\[ \text{Moles of BaCl₂•2H₂O} = \frac{\text{Mass of BaCl₂•2H₂O}}{\text{Molar Mass of BaCl₂•2H₂O}} = \frac{0.437\, \text{g}}{244.27\, \text{g/mol}} \]
2. Calculate the number of moles of Na₂SO₄:
\[ \text{Moles of Na₂SO₄} = \frac{\text{Mass of Na₂SO₄}}{\text{Molar Mass of Na₂SO₄}} = \frac{0.284\, \text{g}}{142.04\, \text{g/mol}} \]
3. Determine the limiting reagent by comparing the mole ratio between BaCl₂ and Na₂SO₄.
4. Use the moles of the limiting reagent to calculate the theoretical yield of BaSO₄.
5. The reaction equation:
\[ \text{BaCl₂ + Na₂SO₄ → BaSO₄ + 2 NaCl} \]
6. Calculate the mass of BaSO₄ precipitate:
\[ \text{Mass of BaSO₄} = \text{Moles of BaSO₄} \times \text{Molar Mass of BaSO₄} \]
By following these steps, you will determine the theoretical yield of BaSO₄ precipitate in grams.
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