A student in an organic chemistry lab prepared di banal acetone according to the

Question

A student in an organic chemistry lab prepared di banal acetone according to the following chemical reaction using excess haze. It is important that the molar ratio of the reactants in this reaction be 2:1. If the student started with 10.6 g of Benz aldehyde and 2.9 g of acetone. what is the theoretical yield of the product? If 10.5 g of the yellow product is obtained, calculi the percent yield. What would be the expected product for a 1:1 mole ratio of the reactants? Another student carries out this reaction using 16 moles of bonze laden de and 10 moles of acetone and obtains 6 moles of dicey acetone. What is the limiting reactant? What is the excess reactant and ho* many moles of this reactant will be left at the end of the reaction? What is the percent yield of this reaction'* iv) What is the actual grams of the yellow product obtained in has reaction '

Explanation / Answer

We have:

Molecular weight of benzaldehyde = 106.23 g/mol

Weight of benzaldehyde (initial) = 10.6 gm

Molecular weight of acetone = 58.08 g/mol

Weight of acetone (initial) = 2.9 gm

Molecular weight of product = 234.30 g/mol

Weight of product obtained (observed yield) = 10.5 gm

Molar ratio of the given reaction = 2:1

To calculate:

Solution:

The reaction is:

2 benzaldehyde + acetone à dibenzalacetone

i.e 2 moles of benzaldehyde and 1 mole of acetone give 1 mole of dibenzalacetone.

a) Theoretical yield-

Now we calculate the number of moles produced when each reactant is used up totally;

1 mol of benzaldeyde = 106.23 g/mol

Then (y) moles = 10.6 gm

Thus y = 10.6*1/106.23 = 0.0993 moles

2 moles of benzaldehyde produce 1 mole of dibenzalacetone

Then 0.0993 moles will produce (q) moles

Thus q = 0.0993*1/2 =0.0496 moles of dibenzalcetone

i.e. in short:

10.6 g Benzaldehyde * 1 mol / 106.23 g/mol * 1 mol dibenzalacetone / 2 mol benzaldehyde

Considering only numbers:

10.6 * 1/106.23 * 1/2 = 0.0496 moles

Similarly calculating for acetone:

2.9 gm of acetone * 1 / 58.08 g/mol * 1 mol dibenzalacetone / 1 mol acetone

Considering only numbers and calculating : 2.9*1/58.08 *1/1 = 0.0499 moles

For calculating the theoretical yield we use the smallest number of moles produced.

In this case: Theoretical yield = 0.0496 moles = 0.0496 * 234.30 = 11.6 gm

b) Percentage yield-

Percentage yield = observed yield / theoretical yield * 100 = 10.5 / 11.6 * 100 = 90.5%

Percentage yield = 90.5%

c) 1:1 ratio

If the molar ratio of the reactant is 1:1

Then the amount of product produced will be :

2 mole of benzaldehyde produces 1 mol of dibenzalacetone

Thus 1 mole will produce ½ mole of product i.e.

The theoretical yield will be halved = 11.6 / 2 = 5.8 gm

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