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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..

  • science(chem) -

    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

  • science(chem) -

    Oh...alright then..

    Thanks Dr.Bob =)

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