I was told before that it takes 2 mol of fluorene to be reduced by 1mol of sodium borohydride....

How many mg of fluorenone can 0.53 mol of sodium borohydride(NaBH4) reduce?

what I got...

0.53mol NaBH4 ( 2mol fluorenone/ 1mol NaBH4)(180.192g/1mol fluorenone)(1000mg/1g)= 19,1003.52mg of fluorenone

Is it just me or does this look really huge in terms of the mg...

well 0.53mol NaBH4 (37.83g/mol)(1000mg/1g)= 20,049.9mg ...used

I don't really know...

Help please..

But 0.53 mol x 2 = 1.06 mols fluorenone and that x 180.12 (a large number) gives about 190 g. Yes, 191,000 mg sounds large but 1 mol of fluorenone is 180,000 mg

Oh...alright then..

Thanks Dr.Bob =)

To determine the amount of fluorenone that can be reduced by 0.53 mol of sodium borohydride (NaBH4), you can use the given ratio of 2 mol of fluorene to 1 mol of NaBH4 as follows:

First, multiply the number of moles of NaBH4 by the conversion factor of (2 mol fluorenone/1 mol NaBH4):

0.53 mol NaBH4 * (2 mol fluorenone/1 mol NaBH4) = 1.06 mol fluorenone

Next, convert moles of fluorenone to milligrams by using the molar mass of fluorenone, which is 180.192 g/mol, and the conversion factor of (1000 mg/1 g):

1.06 mol fluorenone * (180.192 g/1 mol fluorenone) * (1000 mg/1 g) = 191036.032 mg fluorenone

So, based on the calculations, 0.53 mol of NaBH4 can reduce approximately 191,036.032 mg (or 191.036 g) of fluorenone.

Therefore, your initial calculation of 19,1003.52 mg of fluorenone is incorrect due to a misplaced decimal point. The correct value is 191,036.032 mg.

Furthermore, your calculation of 20,049.9 mg of NaBH4 used is also incorrect. You multiplied the given amount of NaBH4 in grams by the conversion factor of (1000 mg/1 g), which is not necessary. Instead, simply use the given value of 0.53 mol NaBH4 in subsequent calculations.