How many mg of fluorenone can 0.53mol of sodium borohyride(NaBH4) reduce?
~I don't know how to get this..wouldn't I have to have a reaction equation given to me??
Well I didn't have one for the question given to me so I don't know what to do. I tried looking for the reaction online but came up empty handed unfortunately.
Could someone help me out on this?
Thank You =)
If you are breaking one = per fluorene, then it takes two hydrogens to do that, from the borohydride.
So for each mole of borohydride, you can break reduce two fluorenone molecules of the double bond.
Thank You Bob
To determine the number of milligrams of fluorenone that can be reduced by 0.53 mol of sodium borohydride (NaBH4), we need to consider the reaction between these two compounds. Unfortunately, since you haven't been provided with a reaction equation, it becomes difficult to provide an exact answer.
Generally, NaBH4 is a reducing agent commonly used in organic chemistry to convert carbonyl compounds (such as fluorenone) into their respective alcohols. The reaction typically proceeds as follows:
NaBH4 + R-C=O → R-CH2OH + NaBO2
Where R-C=O represents the carbonyl compound (fluorenone in this case) and R-CH2OH represents the corresponding alcohol product. NaBO2 is the byproduct of the reaction.
The molar ratio between NaBH4 and fluorenone can be determined from the balanced stoichiometric coefficients in the reaction equation. However, since we don't have a specific equation, we won't be able to calculate an accurate value.
If you are working on a specific experiment or problem that involves this reaction, it would be best to refer to the given information or consult with your instructor or class materials for the required equation or additional information.