Question

hen doing part the above aldol reaction, how many grams of acetone would you need if you are trying to make a maximum theoretical yield of 100 grams of 2,2,8,8-tetramethylnona-3,6-dien-5-one? (12 points) After figuring how many grams of acetone, convert to mL of acetone. (3 points) Show all work and calculations for full credit

Answer 1

29.9g acetone

37.8mL acetone

Step-by-step explanation:

The reaction is:

acetone + 2 2.2-dimethylpropanal → 2,2,8,8-tetramethylnona-3,6-dien-5-one

Acetone (0.791g/mL, 58.08g/mol); 2,2,8,8-tetramethylnona-3,6-dien-5-one (194.13g/mol)

100g of product are:

100g×(1mol/194.13g) = 0.515mol 2,2,8,8-tetramethylnona-3,6-dien-5-one

As 1 mole of product comes from 1 mole of acetone, moles of acetone are 0.515mol. In grams:

0.515mol acetone × (58.08g/mol) = 29.9g acetone

Now, in mL are:

29.9g acetone × (1mL / 0.791g) = 37.8mL acetone