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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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Now we have to calculate the mole fraction of CH_4,CO_2 and He gases.

\text{Mole fraction of }CH_4=\frac{\text{Moles of }CH_4}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}

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and,

\text{Mole fraction of }CO_2=\frac{\text{Moles of }CO_2}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}

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\text{Mole fraction of }He=\frac{\text{Moles of }He}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}

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Now we have to calculate the partial pressure of CH_4,CO_2 and He gases.

According to the Raoult's law,

p_i=X_i\times p_T

where,

p_i = partial pressure of gas

p_T = total pressure of gas  = 5.78 atm

X_i = mole fraction of gas

p_{CH_4}=X_{CH_4}\times p_T

p_{CH_4}=0.267\times 5.78atm=1.54atm

and,

p_{CO_2}=X_{CO_2}\times p_T

p_{CO_2}=0.179\times 5.78atm=1.03atm

and,

p_{He}=X_{He}\times p_T

p_{He}=0.554\times 5.78atm=3.20atm

Thus, the partial pressure of CH_4,CO_2 and He gases are, 1.54, 1.03 and 3.20 atm respectively.

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