Question

A reduction reaction you will see later on is: NaBH4 +4H2O + 4(CH3)2CO -> B(OH3) +...

A reduction reaction you will see later on is:

NaBH4 +4H2O + 4(CH3)2CO -> B(OH3) + NaOH + 4C3H8O

You begin the reaction with 100 mg of acetone and want to use 1.3 equivalents of NaBH4. What mass of NaBH4 should you use?

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Answer #1

1.3 mol NaBH4 reacts with 4 mol acetone as per your requirement

100 mg = 0.1 g acetone; acetone molar mass = 58.08 g/mol

moles acetone = 0.1g / 58.08 g/mol = 0.00172176308 mol, so

moles NaBH4 required = 0.00172176308 x (1.3 / 4 ) = 0.000559573 mol

convert moles NaBH4 to mass NaBH4 by multiplying with it's molar mass

mass of NaBH4 = 0.000559573 mol x 37.83 g/mol = 0.02116864669 g = 21.1686466942 mg ~ 21.2 mg

mass of NaBH4 = 21.2 mg

Hope this helped you!

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