Question

Determine the limiting reagent and calculate the theoretical yield if 0.96 g of benzyltriphenylphosphonium chloride is reacted with 0.30 mL of cinnamaldehyde.  

Answer 1

1 mole of benzyltriphenylphosphonium chloride is reacted with 1 mole of cinnamaldehyde to give 1 mole of 1,4-diphenyl-1,3-butadiene

388.86 g/mol of benzyltriphenylphosphonium chloride is reacted with 132.16 g/mol of cinnamaldehyde to give 206.28 g/mol of 1,4-diphenyl-1,3-butadiene

No. Moles of benzyltriphenylphosphonium chloride = Given weight/molecular weight

                                                                                              = 0.96 (g)/388.86 (g/mol)

                                                                                              = 0.00245 mol

Density of cinnamaldehyde = 1.05 g/cm³

No. of moles of cinnamaldehyde = Given weight/molecular weight

                                                            = (0.30 mL x 1.05 g/mL)/ 132.16 g/mol       {mass= density x volume}

                                                           = 0.00238 mol

  

0.00245 mol of benzyltriphenylphosphonium chloride is reacted with 0.00238 mol of cinnamaldehyde

Hence, the Number of moles of Cinnamaldehyde is less, so it is limiting reagent

132.16 g/mol of cinnamaldehyde gives 206.28 g/mol of 1,4-diphenyl-1,3-butadiene

1 g of cinnamaldehyde gives 206.28 (g/mol)/132.16 (g/mol) of 1,4-diphenyl-1,3-butadiene

0.315 g of cinnamaldehyde gives 206.28 (g/mol)/132.16 (g/mol) x 0.315 g of 1,4-diphenyl-1,3-butadiene

0.315 g of cinnamaldehyde gives 0.492 g of 1,4-diphenyl-1,3-butadiene

Theoretical yield = 0.492 g of 1,4-diphenyl-1,3-butadiene