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bonufazy [111]
3 years ago
13

hen doing part the above aldol reaction, how many grams of acetone would you need if you are trying to make a maximum theoretica

l 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
Chemistry
1 answer:
Elan Coil [88]3 years ago
7 0

Answer:

29.9g acetone

37.8mL acetone

Explanation:

The reaction is:

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

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

<em />

100g of product are:

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

<em />

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) = <em>29.9g acetone</em>

Now, in mL are:

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

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