Question

calculate the weight of sodium borohydride needed to reduce 3.4 grams of benzil to hydrobenzoin?

Answer 1

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

3.4 mg benzil MW = 210.23 g/mole

3.4 mg = 3.4 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