Top topic this week
Back to FAQs
Related FAQs
Frequently Asked Question
Questions asked by users might not always be phrased in the clearest way.
How much sodium borohydride is needed to reduce 75 mg of benzil.
Accepted Answer
To determine the amount of sodium borohydride (NaBH4) needed to reduce 75 mg of benzil, we need to consider the stoichiometry of the reaction. Here's a breakdown:
1. Balanced Chemical Equation:
Benzil (C14H10O2) + 2 NaBH4 → 1,2-Diphenylethane-1,2-diol (C14H14O2) + 2 NaBO2 + 4 H2
2. Molar Masses:
Benzil: 210.23 g/mol
NaBH4: 37.83 g/mol
3. Stoichiometry: The equation shows that 2 moles of NaBH4 are required to reduce 1 mole of benzil.
4. Calculations:
Convert 75 mg of benzil to grams: 75 mg / 1000 mg/g = 0.075 g
Calculate moles of benzil: 0.075 g / 210.23 g/mol = 3.57 x 10^-4 mol
Moles of NaBH4 needed: 3.57 x 10^-4 mol benzil
(2 mol NaBH4 / 1 mol benzil) = 7.14 x 10^-4 mol NaBH4
Mass of NaBH4 needed: 7.14 x 10^-4 mol
37.83 g/mol = 0.027 g = 27 mg
Therefore, approximately 27 mg of sodium borohydride is needed to reduce 75 mg of benzil.
Articles you might like