Answer : The correct answer for theoretical yield = 2.73 g
Theoretical yield : It is defined as maximum amount of yield that is expected to be produce from reactants . The theoretical yield is decided by amount of product formed by Limiting Reagent .
Theoretical yield can be calculated bu using stroichiometric relation between mole of limiting reactant and product .
The reaction of Benzil can be written as :
one mole of Benzoin produces one mole of Benzil ( Image ) .
Since only one reactant , Benzoin is present so limiting reactant can be considered as Benzoin .
Theoretical yield can be calculates in following steps :
Step 1 : Conversion of mass of Benzoin to its mole .
Mass can be converted to mole using following formula :
[tex] Mole = \frac{Mass (g)}{Molar mass \frac{g}{1 mol}} [/tex]
Molar mass of Benzoin (C₁₄H₁₂O₂ ) = 212.24 [tex] \frac{g}{mol} [/tex]
Plugging value of Mass and molar in mole formula :
[tex] Mole = \frac{2.72 g}{212.24 \frac{g}{mol}} [/tex]
Mole = 0.013 mol
Step 2 : To find mole ratio of Benzil to Benzoin .
Mole ratio can be find out from balanced reaction . From balanced reaction one mole of Benzoin giving one mole of Benzil , Hence mole ratio = 1 : 1
Step 3 : Find mole of Benzil
[tex] Mole of Benzil = 0.013 mol of Benzoin * \frac{1 mole of Benzil }{1 mole of Benzoin} [/tex]
Mole of Benzil = 0.013 mol
Step 4 : To find mass of Benzil
Molar mass of Benzil (C₁₄H₁₀O₂ ) = 210.232 [tex] \frac{g}{mol} [/tex]
Plugging value in mole formula
[tex] 0.013 mole of Benzil = \frac{mass of Benzil }{210.232 \frac{g}{mol}} [/tex]
Multiplying both side by 210.232 [tex] \frac{g}{mol} [/tex]
[tex] 0.013 mol * 210.232 \frac{g}{mol} = \frac{mass of benzil}{210.232 \frac{g}{mol}} * 210.232 \frac{g}{mol} [/tex]
Mass of Benzil = 2.73 g
Hence Theoretical mass of Benzil = 2.73 g
