Question

How to calculate the volume of acetophenone that is not given????

when adding 3.0 ml of P-anisaldehyde and an equimolar amount of acetophenone in a flask. which is the limiting reagent in aldol condensation reaction?

Answer 1

Ans. Part 1: Given, volume of p-anisaldehyde = 3.0 mL

Now,

Mass of p-anisaldehyde = volume x density = 3.0 mL x (1.119 g/ mL) = 3.357 g

Moles of p-anisaldehyde = Mass/ molar mass

= 3.357 g / (136.15 g/mol)

= 0.024657 mol

Hence, moles of p-anisaldehyde taken = 0.024657 mol

Part 2: Given, Amount of acetophenone taken = Equimolar amount of p-anisaldehyde

The terms “equimolar” means “equal number of moles of two chemical species”.

Therefore,

            Moles of acetophenone taken = Moles of p-anisaldehyde taken

            Hence, Moles of acetophenone taken = 0.024657 mol

# Now,

            Mass of acetophenone taken = 0.024657 mol x (120.15 g/ mol) = 2.963 g

And,

            Volume of acetophenone taken = Mass / density

                                                            = 2.963 g / (1.03 g/ mL)

                                                            = 2.88 mL

Therefore, volume of acetophenone taken = 2.88 mL

# Note: Determine the limiting reactant depend on the reaction given to you.