Question

The reaction is run using 0.20 g of anisole with 2.2 molar equivalents of AlCl3 and 1.1 molar equivalents of acetic anhydride.How to find mass of anisol and aluminium chloride and volume of acetic anhydride to be used.

Answer 1

Given the mass of anisle(C6H5OCH3) = 0.20 g

Molecular mass of anisole = 108.1 g/mol

Hence moles of anisle taken = mass / molar mass = 0.20 g / 108.1 g/mol = 0.00185 mol anisole

2.2 molar equivalents of AlCl3 =  0.00185 mol x 2.2 = 0.00407 mol AlCl3

Molecular mass of AlCl3 = 133.34 g/mol

Hence mass of AlCl3 used = 0.00407 mol AlCl3 x 133.34 g/mol = 0.543 g (answer)

1.1 molar equivalents of acetic anhydride(C4H6O3) =  0.00185 mol x 1.1 = 0.00185 mol AlCl3

Molecular mass of acetic anhydride(C4H6O3) = 102.09 g/mol

Hence mass of acetic anhydride(C4H6O3) used = 0.00185 mol C4H6O3 x 102.09 g/mol = 0.189 g

The volume of acetic anhydride need can be calculated from the concentration of acetic acid