3CH3COOH+2C3H5OH=2CH3CooC3H5+4H2O

Chemistry
Tutor: None Selected Time limit: 1 Day

A. What mass of propanol (C3H5OH) must be available to completely consume 6.70 moles of acetic acid (CH3COOH)?

B. What would be the theoretical yield (in grams) of propyl acetate (CH3COOC3H5)?

Apr 22nd, 2015

Molecular mass of propanol = 60.09g/mol

A) According to balanced reaction,

3 moles of acetic acid is required to consume 2 moles of propanol

Hence for one mole of acetic acid propanol required = 2/3 moles

hence for 6.7 moles of acetic acid propanol required = 2/3 *6.7 moles

Hence total mass of propanol requirred = molecular mass*number of moles of propanol

                                                               = 6.7*60.09*2/3

                                                                =268.40 g

B) Number of moles of propyl acetate generated = 2/3*number of moles of acetic acid consumed = 2/3*6.70 moles

Molecular mass of propyl acetate = 102.81 g/mol

Hence total theoretical yield in grams of propyl acetate = molecular mass*moles

   = 6.7*102.81*2/3

=459.21 g

Apr 22nd, 2015
I hope you are satisfied with my detailing. Please give a good rating if you are happy. Thank you 
Apr 22nd, 2015

can you explain how you found 6.7?

Apr 22nd, 2015

Hey it is given as a initial number of moles of acetic acid. It is given in question, in fact with that information only we are able to calculate all other quantities.

In any chemical reaction you at least need one known quantity to calculate the rest of it.

I hope it helps. 

Apr 22nd, 2015

sorry, i should have realized.. thank you!

Apr 22nd, 2015

Thanks for adding me as tutor feel free to ask me any questions you want. If I am available I will certainly answer you.

Apr 22nd, 2015

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