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Calculate the theoretical weight of sodium borohydride needed to reduce 75mg of benzil?

Accepted Answer

To calculate the theoretical weight of sodium borohydride (NaBH4) needed to reduce 75 mg of benzil, we need to consider the stoichiometry of the reaction and the molar masses of the compounds involved. Here's a breakdown:
1. Balanced Chemical Equation:
Benzil (C14H10O2) + 2 NaBH4 → 1,2-Diphenylethanediol (C14H14O2) + 2 NaBO2 + 2 H2
2. Molar Masses:
Benzil (C14H10O2): 210.23 g/mol
Sodium Borohydride (NaBH4): 37.83 g/mol
3. Stoichiometry:
From the balanced equation, 2 moles of NaBH4 are required to reduce 1 mole of benzil.
4. Calculations:
Convert the mass of benzil to moles: (75 mg / 1000 mg/g) / (210.23 g/mol) = 0.000357 moles of benzil
Calculate the moles of NaBH4 needed: 0.000357 moles benzil
(2 moles NaBH4 / 1 mole benzil) = 0.000714 moles NaBH4
Convert the moles of NaBH4 to grams: 0.000714 moles NaBH4
(37.83 g/mol) = 0.0270 g NaBH4
Therefore, you would theoretically need approximately 27 mg of sodium borohydride to reduce 75 mg of benzil.

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