<u>Answer:</u> The theoretical yield of triphenylmethanol is 0.173 grams
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
.....(1)
Given mass of benzophenone = 121 mg = 0.121 g (Conversion factor: 1 g = 1000 mg)
Molar mass of benzophenone = 182.21 g/mol
Putting values in equation 1, we get:
![\text{Moles of benzophenone}=\frac{0.121g}{182.21g/mol}=6.64\times 10^{-4}mol](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20benzophenone%7D%3D%5Cfrac%7B0.121g%7D%7B182.21g%2Fmol%7D%3D6.64%5Ctimes%2010%5E%7B-4%7Dmol)
The chemical equation for the formation of triphenylmethanol from benzophenone follows:
![\text{Benzophenone}+Mg+\text{Bromobenzene}\rightarrow \text{Triphenylmethanol}+MgBr](https://tex.z-dn.net/?f=%5Ctext%7BBenzophenone%7D%2BMg%2B%5Ctext%7BBromobenzene%7D%5Crightarrow%20%5Ctext%7BTriphenylmethanol%7D%2BMgBr)
As, bromobenzene is the limiting reagent. So, it will limit the formation of products
By Stoichiometry of the reaction:
1 mole of bromobenzene produces 1 mole of triphenylmethanol
So,
of bromobenzene will produce =
moles of triphenylmethanol
Now, calculating the mass of triphenylmethanol from equation 1, we get:
Molar mass of triphenylmethanol = 260.33 g/mol
Moles of triphenylmethanol =
moles
Putting values in equation 1, we get:
![6.64\times 10^{-4}mol=\frac{\text{Mass of triphenylmethanol}}{260.33g/mol}\\\\\text{Mass of triphenylmethanol}=(6.64\times 10^{-4}mol\times 260.33g/mol)=0.173g](https://tex.z-dn.net/?f=6.64%5Ctimes%2010%5E%7B-4%7Dmol%3D%5Cfrac%7B%5Ctext%7BMass%20of%20triphenylmethanol%7D%7D%7B260.33g%2Fmol%7D%5C%5C%5C%5C%5Ctext%7BMass%20of%20triphenylmethanol%7D%3D%286.64%5Ctimes%2010%5E%7B-4%7Dmol%5Ctimes%20260.33g%2Fmol%29%3D0.173g)
Hence, the theoretical yield of triphenylmethanol is 0.173 grams