Answer:
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.
<h2>So, in conclusion,
Limiting reagent NaBH₄</h2>
Hope this helps
