Calculate the theoretical yield in grams of brominated limonene if you reacted 0.250 grams of limonene with excess bromine solution (this means that limonene is the limiting reagent).

Question

Limonene MW= 136 g/mol & Brominated Limonene MW=455.9 g/mol

Answer 1

At first this question can seem a little confusing since the reaction's equation isn't specified. However, in these inorganic reactions, bromine tends to simply substitute hydrogens. Therefore, let's assume that each mole of Limonene forms one mole of Brominated Limonene. If you do the math using the two supplied molecular weights in addition to the molecular weight of bromine (79.9 g/mol), you'll discover that each mole of Brominated Limonene has 4 Br atoms.

First, let's calculate the number of moles of Limonene:

n = m / MW = 0.250 g / 136 g/mol = 0.001838 mol

Since the number of moles of Brominated Limonene is equal to the number of moles of Limonene, we know that there are also 0.001838 mol of Brominated Limonene. Now let's calculate the expected mass yield of Brominated Limonene:

m = n * MW = 0.001838 mol * 455.9 g/mol = 0.838 g