[Aznhmonglor]

How many moles of a cyclohexanone could be reduced with 3.8g of NaBH4?

On this question I know that cyclohexanone has a mass of 98.15 g/mol. Im not sure how I am to start this question.

[azmanam]

start with a balanced equation

[Aznhmonglor]

hmm not sure how to begin that but what does cyclohexanone and NaBH4 make?

is the equation this?

C6H10O + NaBH4 -> C6H10O + NaBH2

[macman104]

You must have a book that explains reduction of ketones, right?

[Morderkerl]

The first thing to realize is that one mole of NaBH4 provides 1 mole of hydride. Which means if you have 3.8 g of NaBH4, with a molecular weight of 37.83, tou have 0.1 moles of NaBH4 and 0.1 moles of Hydride. In order to reduce cyclohexanone you also need one mole of Hydride for ever mole of ketone. Thus if you multiply the 0.1 mole with the molar mass of cyclohexanone (98g/mole) you get 9.84g. Thus with 3,8g of NaBH4 you can reduce 9.84g cyclohexanone.

[macman104]

Sorry, Morderkerl, but I disagree.  One mole of NaBH4 has a total of 4 hydrides it can provide.  It may not work to 100% efficiency, but it can reduce 4 times.

[azmanam]

one mole of NaBH4 provides 1 mole of hydride

Nope.  this is the key to the problem.  1 mol of NaBH₄ provides four mols of hydride.  Soichiometry is 1:4.

http://www.cem.msu.edu/~reusch/VirtTxtJml/aldket1.htm#rx2a

[Morderkerl]

Sorry, Morderkerl, but I disagree.  One mole of NaBH4 has a total of 4 hydrides it can provide.  It may not work to 100% efficiency, but it can reduce 4 times.

Yes it can provide 4 hydrides, but as soon as the first hydride leaves the extend of hydride production reduces. If this is then the case, to come back to the original question, you can reduce 0.4 moles of cyclohexanone, because each mole of NaBH4 can provide 4 moles of hydride, you need one mole of hydride to reduce one mole of ketone (the other hydrogen comes from water) and since there are 0.1 moles of NaBH4 you can reduce four times that.