Question

What would be the theoretical yield of the hydrogenation of castor oil?

- 1.0 g castor oil

- 1.3 g limonene

- 0.1 g 5% Pd/C

Answer 1

Molar mass of castor oil = 933.40 g/mol

Number of moles of castor oil = (1.0 g)/(933.40 g/mol) = 1.071352 milli moles

Molar mass of limonene = 136.24 g/mol

Number of moles of limonene = (1.3 g)/(136.24 g/mol) = 9.542 milli moles

1 mole castor oil + 3 moles limonene ---------> 1 mole hydrogenated castor oil

Therefore we get 1.071352 milli moles of castor oil is the limiting reactant. It reacts with 3.214056 milli moles of limonene. Produces 1.071352 milli moles of product.

Pd/C is the catalyst here

Molar mass of product = 936.42352 g/mol

Theoretical yield= 0.001071352 mol X 936.42352 g/mol = 1.003239 grams.

Answer: 1.003239 grams