Question

If you started with 0.274g of benzil and 0.198g of dibenzyl ketone:

1) What is the limiting reagent? How do we find out which one is limiting reagent?

2) What is the theoretical yield of tetraphenylcyclopentadienone? Please show detailed caclulations.

Thank You!

Answer 1

1) Bezil + dibenzyl ketone Tetraphenylcyclopentadienone

molar mass of Bezil = 210.232 gm/mol that mean 1 mole of Bezil = 210.232 gm

molar mass of dibenzyl ketone = 210.28 gm/mol that mean 1 mole of dibenzyl ketone = 210.28 gm

molar mass of Tetraphenylcyclopentadienone = 374.478 gm/mol that mean 1 mole of Tetraphenylcyclopentadienone = 384.478 gm

1 mole of bezil react with 1 mole of dibenzyl ketone that mean 210.232 gm of bezil react with 210.28 gm of dibenzyl ketone then to react with 0.274 gm of bezil required = 210.28 0.274 / 210.232 = 0.27406256 gim of dibenzyl ketone but dibenzyl ketone given only 0.198 gm thus dibenzyl ketone is the limiting reagent

2) dibenzyl ketone is the limiting reagent react completly 1 mole of dibenzyl ketone produce 1 mole of Tetraphenylcyclopentadienone that mean 210.28 gm of dibenzyl ketone produce 374.478 gm of Tetraphenylcyclopentadienone then 0.198 gm of dibenzyl ketone produce = 0.198 374.478/210.28 = 0.3526 gm of Tetraphenylcyclopentadienone.

therotical yield =0.3526 gm of Tetraphenylcyclopentadienone