Calculate the theoretical weight of sodiumborohydride needed to reduce 50 mg of

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Explanation / Answer

2CR2CO(aq) + 2H+(aq) + 2e - --> 2CR2COH(aq) (* 2 for since 2 ketones for benzil) * 2 BH4-(aq) + 3H2O(l) --> BO3-3(aq) + 7H+(aq) + 4e- add 2CR2CO(aq) + NaBH4(aq) + 4H2O(l) --> 2CR2COH(aq) + H3BO3(aq) so there will be a 2:1 mole equivalence between benzil and the NaBH4 50 mg benzil MW = 210.23 g/mole 50 mg = 50 mg * mole/210.23g * g/1000mg = 2.4 * 10^-4 mole 1/2 mole NaBH4 will be required = 1.2* 10-4 mole NaBH4 MW = 37.83 g/mole 1.2 * 10^-4 mole NaBH4 = 1.2 * 10^-4 mole * 37.83 g/mole = = 0.0045 g NaBH4 = 4.5 mg NaBH4

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