How many grams of Aldol product can be produced from the complete reaction of 0.2 grams of vanillin with an excess of acetone in the presence of aqueous base? Enter only the number with two significant figures.

There is 0.25 grams of C11H12O3 produced

Explanation:

Step1: Data given

vanillin = C8H8O3

Mass of vanillin = 0.2 grams

Molar mass of vanillin = 152.15 g/mol

Acetone = 58.08 g/mol

Step 2: The balanced equation

C8H8O3 + C3H6O → H2O + C11H12O3

Step 3: Calculate moles of C8H8O3

Moles C8H8O3 = mass / molar mass

Moles C8H8O3 = 0.2 grams / 152.15 g/mol

Moles C8H8O3 = 0.0013 moles

Step 4: Calculate moles of C11H12O3

For 1 mol vanillin we need 1 mol acetone to produce 1 mol C11H12O3

Step 5: Calculate mass of C11H12O3

Mass C11H12O3 = moles * molar mass

Mass C11H12O3 = 0.0013 moles * 192.21 g/mol

Mass C11H12O3 = 0.25 grams

There is 0.25 grams of C11H12O3 produced