For the reaction shown below, 325 mg of benzaldehyde, 90 mg of acetone, and excess sodium hydroxide are used to obtain 66.1 mg of dibenzalacetone. What is the percent yield of dibenzalacetone? NaOH benzaldehyde MW = 106 g/mol; acetone MW = 58.0 g/mol; sodium hydroxide MW = 40.0 g/mol; dibenzalacetone MW = 234 g/mol
For the reaction shown below, 325 mg of benzaldehyde, 90 mg of acetone, and excess sodium hydroxide are used to obtain 66.1 mg of dibenzalacetone. What is the percent yield of dibenzalacetone? NaOH benzaldehyde MW = 106 g/mol; acetone MW = 58.0 g/mol; sodium hydroxide MW = 40.0 g/mol; dibenzalacetone MW = 234 g/mol
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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![**Calculating Percent Yield of Dibenzalacetone**
For the reaction shown below, 325 mg of benzaldehyde, 90 mg of acetone, and excess sodium hydroxide are used to obtain 66.1 mg of dibenzalacetone. To determine the percent yield of dibenzalacetone, we will follow these steps:
1. **Identify the Molar Masses:**
- Benzaldehyde (C₇H₆O): MW = 106 g/mol
- Acetone (C₃H₆O): MW = 58.0 g/mol
- Sodium Hydroxide (NaOH): MW = 40.0 g/mol (though not needed for yield calculation)
- Dibenzalacetone (C₁₇H₁₄O): MW = 234 g/mol
2. **Write the balanced chemical equation:**
![Chemical Reaction]
Benzaldehyde (C₇H₆O) + Acetone (C₃H₆O) + NaOH → Dibenzalacetone (C₁₇H₁₄O)
3. **Calculate moles of reactants:**
- Moles of Benzaldehyde = \( \frac{325 \text{ mg}}{106 \text{ g/mol}} = \frac{325 \times 10^{-3} \text{ g}}{106 \text{ g/mol}} \approx 3.07 \times 10^{-3} \text{ mol} \)
- Moles of Acetone = \( \frac{90 \text{ mg}}{58.0 \text{ g/mol}} = \frac{90 \times 10^{-3} \text{ g}}{58.0 \text{ g/mol}} \approx 1.55 \times 10^{-3} \text{ mol} \)
4. **Determine the limiting reagent:**
The reaction needs 2 moles of benzaldehyde for every 1 mole of acetone.
- Required moles of Benzaldehyde for 1.55 \(\times\) 10\(^{}\)-3 mol Acetone = \( 1.55 \times 10^{-3} \text{ mol} \times 2 = 3.10 \times 10^{-3} \text{](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcadcb35e-d490-4595-9943-db4a09757986%2Fbcbe8e58-081d-4ac9-baa2-c43df63cbb5a%2Fm3l3e3h_processed.png&w=3840&q=75)
Transcribed Image Text:**Calculating Percent Yield of Dibenzalacetone**
For the reaction shown below, 325 mg of benzaldehyde, 90 mg of acetone, and excess sodium hydroxide are used to obtain 66.1 mg of dibenzalacetone. To determine the percent yield of dibenzalacetone, we will follow these steps:
1. **Identify the Molar Masses:**
- Benzaldehyde (C₇H₆O): MW = 106 g/mol
- Acetone (C₃H₆O): MW = 58.0 g/mol
- Sodium Hydroxide (NaOH): MW = 40.0 g/mol (though not needed for yield calculation)
- Dibenzalacetone (C₁₇H₁₄O): MW = 234 g/mol
2. **Write the balanced chemical equation:**
![Chemical Reaction]
Benzaldehyde (C₇H₆O) + Acetone (C₃H₆O) + NaOH → Dibenzalacetone (C₁₇H₁₄O)
3. **Calculate moles of reactants:**
- Moles of Benzaldehyde = \( \frac{325 \text{ mg}}{106 \text{ g/mol}} = \frac{325 \times 10^{-3} \text{ g}}{106 \text{ g/mol}} \approx 3.07 \times 10^{-3} \text{ mol} \)
- Moles of Acetone = \( \frac{90 \text{ mg}}{58.0 \text{ g/mol}} = \frac{90 \times 10^{-3} \text{ g}}{58.0 \text{ g/mol}} \approx 1.55 \times 10^{-3} \text{ mol} \)
4. **Determine the limiting reagent:**
The reaction needs 2 moles of benzaldehyde for every 1 mole of acetone.
- Required moles of Benzaldehyde for 1.55 \(\times\) 10\(^{}\)-3 mol Acetone = \( 1.55 \times 10^{-3} \text{ mol} \times 2 = 3.10 \times 10^{-3} \text{
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