Calculations for benzaldehyde, acetone, and dibenzalacetone benzaldehyde acetone dibenzalacetone Molar mass (g/mol) Mass used or produced Moles (g) 106.12 2.650 .02497 58.08 0.730 .0126 234.29 1.481 .00632 (0/1pts) From te table above, the limiting reagent is: acetone (-1 pts) Incorrect. Take into account both the moles of each reagent and their mole to mole ratio with the product. (0/1pts) Theoretical yield (g) 1.450 (-1 pts) Incorrect. Make sure to determine the theoretical yield based on the limiting reagent. (0/1pts) Percent yield (g) 25.316
The Correct Answer and Explanation is:
Balanced Chemical Equation:
acetone+2 benzaldehyde→dibenzalacetone+2 H2O\text{acetone} + 2 \, \text{benzaldehyde} \rightarrow \text{dibenzalacetone} + 2 \, \text{H}_2\text{O}acetone+2benzaldehyde→dibenzalacetone+2H2O
This shows that:
- 1 mole of acetone reacts with 2 moles of benzaldehyde to form 1 mole of dibenzalacetone.
- Mole ratio: acetone : benzaldehyde = 1 : 2
Step 1: Moles of Reactants
| Substance | Molar Mass (g/mol) | Mass (g) | Moles (mol) |
|---|---|---|---|
| Benzaldehyde | 106.12 | 2.650 | 0.02497 |
| Acetone | 58.08 | 0.730 | 0.01257 |
To react with 0.01257 mol of acetone, we would need: 0.01257×2=0.02514 mol of benzaldehyde0.01257 \times 2 = 0.02514 \, \text{mol of benzaldehyde}0.01257×2=0.02514mol of benzaldehyde
But we only have 0.02497 mol of benzaldehyde, which is slightly less than needed. Therefore:
Limiting Reagent: Benzaldehyde
Step 2: Theoretical Yield of Dibenzalacetone
From the stoichiometry:
- 2 mol benzaldehyde → 1 mol dibenzalacetone
- So, 0.02497 mol benzaldehyde →
0.024972=0.012485 mol dibenzalacetone (theoretical)\frac{0.02497}{2} = 0.012485 \, \text{mol dibenzalacetone (theoretical)}20.02497=0.012485mol dibenzalacetone (theoretical)
Using molar mass of dibenzalacetone (234.29 g/mol): Theoretical yield=0.012485×234.29=2.924 g\text{Theoretical yield} = 0.012485 \times 234.29 = \boxed{2.924 \, \text{g}}Theoretical yield=0.012485×234.29=2.924g
Step 3: Percent Yield
Actual mass of dibenzalacetone produced = 1.481 g
Theoretical yield = 2.924 g Percent yield=(1.4812.924)×100=50.6%\text{Percent yield} = \left(\frac{1.481}{2.924}\right) \times 100 = \boxed{50.6\%}Percent yield=(2.9241.481)×100=50.6%
Final Answers:
- Limiting reagent: Benzaldehyde
- Theoretical yield: 2.924 g
- Percent yield: 50.6%
Summary Explanation:
To identify the limiting reagent in a chemical reaction, the molar amounts of each reactant must be compared using the stoichiometric ratios from the balanced equation. Although acetone had a smaller number of moles, it was not the limiting reagent because it required only half the molar amount of benzaldehyde. Since benzaldehyde was present in slightly less than the needed 2:1 ratio, it limited the amount of product that could form. From this, the theoretical yield was calculated using molar relationships, and the percent yield was found by comparing actual to theoretical yields.
