Name Mass Formula Mass moles Draw v Calculations T Read aloud Ba(OH)₂ + 2Na₂SO4 → BaSO+2NaOH 1200g 171.34 g/mol 2500g 142.04 g/mol A 233.39 g/mol 40.00 g/mol

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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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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### Chemical Reaction: Barium Hydroxide and Sodium Sulfate

In this experiment, we observe the chemical reaction between barium hydroxide and sodium sulfate:

\[ \text{Ba(OH)}_2 + 2\text{Na}_2\text{SO}_4 \rightarrow \text{BaSO}_4 + 2\text{NaOH} \]

#### Reactants and Products Information

1. **Ba(OH)₂**:
   - **Mass**: 1200 g
   - **Formula Mass**: 171.34 g/mol

2. **2Na₂SO₄**:
   - **Mass**: 2500 g
   - **Formula Mass**: 142.04 g/mol

3. **BaSO₄**:
   - **Formula Mass**: 233.39 g/mol

4. **2NaOH**:
   - **Formula Mass**: 40.00 g/mol

#### Moles Calculation
- Use the formula:

  \[
  \text{Moles} = \frac{\text{Mass}}{\text{Formula Mass}}
  \]

Complete calculations based on the formula mass and given mass of each substance.

#### Explanation
This setup helps in understanding stoichiometry by calculating the moles of each reactant and product. The reaction showcases the formation of barium sulfate (a precipitate) and sodium hydroxide from the reaction of barium hydroxide with sodium sulfate.
Transcribed Image Text:### Chemical Reaction: Barium Hydroxide and Sodium Sulfate In this experiment, we observe the chemical reaction between barium hydroxide and sodium sulfate: \[ \text{Ba(OH)}_2 + 2\text{Na}_2\text{SO}_4 \rightarrow \text{BaSO}_4 + 2\text{NaOH} \] #### Reactants and Products Information 1. **Ba(OH)₂**: - **Mass**: 1200 g - **Formula Mass**: 171.34 g/mol 2. **2Na₂SO₄**: - **Mass**: 2500 g - **Formula Mass**: 142.04 g/mol 3. **BaSO₄**: - **Formula Mass**: 233.39 g/mol 4. **2NaOH**: - **Formula Mass**: 40.00 g/mol #### Moles Calculation - Use the formula: \[ \text{Moles} = \frac{\text{Mass}}{\text{Formula Mass}} \] Complete calculations based on the formula mass and given mass of each substance. #### Explanation This setup helps in understanding stoichiometry by calculating the moles of each reactant and product. The reaction showcases the formation of barium sulfate (a precipitate) and sodium hydroxide from the reaction of barium hydroxide with sodium sulfate.
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