Chemistry & Chemical Reactivity
10th Edition
ISBN:9781337399074
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Chapter23: Carbon: Not Just Another Element
Section: Chapter Questions
Problem 66PS
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Transcribed Image Text:1) mass of benzil = moles x molar mass of benzil
= 0.35 mmol x 210.232 g/mol
= 73.5812 x 10^-3 g
= 73.58 mg
mass of benzoin=0.35 mmol x 212.24 g/mol
= 74.284 x 10^-3 g
= 74.284 mg
2)
a) one mole of sodium borohydride (NABH4) can reduce four moles of aldehyde or
ketone
therefore
one mole of sodium borohydride (NaBH4) can reduce 2 moles of benzil because benzil
has 2 ketone group
and
for 0.35 mmol of benzil , moles of NABH4 required = 0.35 mmol / 2 = 0.175 mmol
thus
mass of NaBH4 = 0.175 mmol x 37.83 g/mol
= 6.62 mg
b) as we know that for 0.35 mmol of benzil , only 6.62 mg of NaBH4 is required
therefore
15 mg is very large and Only 0.35 mmol of benzil , leads to the result that
benzil is the limiting reagent.
3) the balanced reaction or the reaction of the NaBH4 with H2O is given as
NABH4 + 3 H20
NaH2BO3 + 4 H2

Transcribed Image Text:A
ABC is triangle with D and E the midpoints of
AB and AC respectively.
E
Prove that BC=2DE.
В
C
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