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What is 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. Here's the breakdown:
1. Balanced Reaction:
Benzil (C14H10O2) + 2 NaBH4 → 2 C14H12O + Na2B2O4 + 4 H2
2. Molar Mass:
Benzil (C14H10O2): 210.23 g/mol
NaBH4: 37.83 g/mol
3. Moles of Benzil:
75 mg benzil × (1 g / 1000 mg) × (1 mol / 210.23 g) = 0.000357 mol benzil
4. Moles of NaBH4:
From the balanced reaction, 1 mol of benzil reacts with 2 moles of NaBH4. Therefore:
0.000357 mol benzil × (2 mol NaBH4 / 1 mol benzil) = 0.000714 mol NaBH4
5. Mass of NaBH4:
0.000714 mol NaBH4 × (37.83 g/mol) = 0.0270 g NaBH4
Therefore, theoretically, you would need approximately 27 mg of sodium borohydride to completely reduce 75 mg of benzil.
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