1) mass of benzil moles x molar mass of benzil 0.35 mmol x 210.232 g/mol 73.5812 x 10^-3 g =73.58 mg
mass of benzoin = 0.35 mmol x 212.24 g/mol 74.284 x 10^-3 g = 74.284 mg
2)
a) one mole of sodium borohydride (NaBH4) can reduce four moles of aldehyde or ketone therefore one mole of sodium borohydride (NaBH4) can reduce 2 moles of benzil because benzil has 2 ketone group and for 0.35 mmol of benzil, moles of NaBH4 required = 0.35 mmol/2=0.175 mmol thus mass of NaBH4 = 0.175 mmol x 37.83 g/mol = 6.62 mg
b) as we know that for 0.35 mmol of benzil, only 6.62 mg of NaBH4 is required therefore 15 mg is very large and Only 0.35 mmol of benzil, leads to the result that benzil is the limiting reagent.
The Correct Answer and Explanation is :
Let’s first break down the calculations and steps:
Part 1:
You provided the mass calculations for benzil and benzoin:
1) Mass of Benzil:
- You calculated the mass of benzil using the formula:
[ \text{Mass} = \text{moles} \times \text{molar mass} ]
For benzil:
[ \text{Mass of benzil} = 0.35 \, \text{mmol} \times 210.232 \, \text{g/mol} = 73.58 \, \text{mg} ]
This is correct.
2) Mass of Benzoin:
- You performed a similar calculation for benzoin:
[ \text{Mass of benzoin} = 0.35 \, \text{mmol} \times 212.24 \, \text{g/mol} = 74.284 \, \text{mg} ]
This is also correct.
Part 2:
a) Calculation of NaBH₄:
- As you correctly pointed out, sodium borohydride (NaBH₄) can reduce 4 moles of aldehyde or ketone, and since benzil contains 2 ketone groups, one mole of NaBH₄ can reduce 2 moles of benzil.
- Given that you have 0.35 mmol of benzil, the moles of NaBH₄ required would be:
[ \text{Moles of NaBH₄} = \frac{0.35 \, \text{mmol}}{2} = 0.175 \, \text{mmol} ] - The mass of NaBH₄ required is:
[ \text{Mass of NaBH₄} = 0.175 \, \text{mmol} \times 37.83 \, \text{g/mol} = 6.62 \, \text{mg} ]
This calculation is correct.
b) Comparison of Benzil and NaBH₄:
- You mentioned that 15 mg of NaBH₄ is excessive compared to the 6.62 mg required. You correctly concluded that since only 6.62 mg of NaBH₄ is needed to reduce 0.35 mmol of benzil, benzil is the limiting reagent in this reaction.
- This conclusion is right because, with 0.35 mmol of benzil, the reaction will use up the available benzil first, leaving excess NaBH₄.
Explanation:
In this reaction, sodium borohydride acts as the reducing agent and benzil is the substance being reduced. The reaction involves the reduction of the two ketone groups in benzil. The amount of NaBH₄ required to reduce benzil is directly related to the number of ketone groups present. Since each mole of NaBH₄ can reduce 2 moles of ketones, the amount of NaBH₄ needed is halved compared to the amount of benzil.
Because you are only working with 0.35 mmol of benzil, you need 0.175 mmol of NaBH₄, which corresponds to 6.62 mg. Given that you have 15 mg of NaBH₄ available, the excess NaBH₄ is not going to be used in the reaction, and the limiting reagent is the benzil.
Image:
Would you like me to generate an image illustrating the reaction between benzil and NaBH₄ or the structure of the molecules involved?