For the reaction of reducing benzil (MW 210.23 g/mol) with sodium borohydride (MW 37.83 g/mol), if 2.56 g of benzil and 0.38 g of sodium borohydride were used to make hydrobenzoin (MW 214.26 g/mol), and 2.22 g of hydrobenzoin was obtained, which chemical is limiting reagent

NaBH₄

Explanation:

First, we need to write the chemical formula of all the compounds:

Benzil: C₁₄H₁₀O₂

Sodium Borhydride: C₁₄H₁₀O₂

Hydrobenzoin: C₁₄H₁₄O₂

Now, let's write the reaction that is taking place and write all the products:

C₁₄H₁₀O₂ + 2NaBH₄ + 2H₂O > C₁₄H₁₄O₂ + 2BH₃ + 2NaOH

We can see that the reaction is already balanced, so we don't need to do anything else.

The question of this exercise is to determine the limiting reagent of the reaction, in other words, the reagent that controls the reaction and produces the 2.22 g of the hydrobenzoin. And to know this we need to see the mole ratio in both reactants, and compare them to the given moles (That can be obtained with the given masses and MW)

According to the above reaction, we have a mole ratio of 1:2, so, let's calculate the moles of benzil and the borohydride, and see which of them is the limiting reactant:

moles C₁₄H₁₀O₂ = 2.56 / 210.23 = 0.0122 moles

moles NaBH₄ = 0.38 / 37.83 = 0.01 moles

moles  C₁₄H₁₄O₂ = 2.22 / 214.26 = 0.0103 moles

We have the moles of every species, now, let's see the mole ratio

If 1 mole of C₁₄H₁₀O₂ > 2 moles of NaBH₄

Then 0.0122 moles C₁₄H₁₀O₂ > X moles of NaBH₄

Solving for X:

X = 0.0122 * 2 / 1 = 0.0244 moles of NaBH₄ are required.

However, we only have 0.01 moles of NaBH₄, and we need so much more of this to completely react with the moles of the benzil. Therefore we can safely assume that the limiting reagent is the NaBH₄

Another data that we can use for this, is the fact the produced moles were 0.0103, and this value is nearest to the moles of NaBH₄ rather than the moles of the benzil.

Hope this helps

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