Question

In a reduction experiment, 45.00 g of acetone, C3H6O, MWt = 58.08, is reduced with sodium borohydride, NaBH4, MWt = 37.83, to form 2-propanol, C3H8O, MWt = 60.10.

Write the balanced equation showing the overall stoichiometry of the reaction. The fate of the borohydride is not necessary, just the mole ratio to the acetone.

How many grams of NaBH4 are required to reduce 45.00 g of acetone?

. If 40.00 gm of product are recovered, what is the percent yield for this reaction?

If acetone is reduced with NaBH4, and workup is completed by adding DCl, (deuterated hydrochloric acid), what is the structure of the resulting product?

1-propanol, C3H8O can be oxidized by NaOCl(aq), to propionic acid, C3H6O2 and NaCl(aq). Balance these half cells using the half cell method. You do NOT need to balance the entire redox reaction. (10 pt) C3H8O ---> C3H6O2 OCl-(aq) ---> Cl-(aq)

Answer 1

Balanced equation:

C₃H₆O + 2 NaBH₄ --> C₃H₇OH + 2 NaBH₃

₁₎ How many grams of NaBH4 are required to reduce 45.00 g of acetone?

Let's turn de 45 g to moles:

moles of acetone = 45 g / 58.08 g/mole = 0.775 moles

now, if 1 mole of acetone reacts with two moles of borohydride, then 0.775 moles...

1 mol acetone ----- 2 moles borohydride

0.775 moles acetone --- x

x= (0.775 * 2) / 1 = 1.55 moles borohydride

grams of borohydride = 1.55 moles * 37.83 g/mole = 58.6 grams

2) If 40.00 gm of product are recovered, what is the percent yield for this reaction?

Let's assume that the limitant reagent is the acetone, so let's calculate the theoretical yield:

1 mole acetone ---- 1 mol of C3H8O

0.775 moles ------- x

x= (0.775 *1) / 1= 0.775 moles C3H8O

theoretical grams = 0.775 moles * 60.10 g/mole = 46.57 g

so, the yield is Y= 40/46.57 * 100% = 86 %

3) can't answer that

4) H2O + C3H8O ---> C3H6O2 + 2H+  + (2e-)

(2e-) + 2H+ + OCl - --> Cl- + H2O

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C2H8O + OCl - --> C3H6O2 + Cl -