What is the actual yield?

Question

Sb+O2→Sb4O6

If an excess amount of Sb reacts with 1.87 g of O₂ during a lab, what is the actual yield, in g, produced?

Answer 1

First, you must balance the equation:

4Sb + 3O₂→Sb₄O₆

Next, since Sb is in excess, the amount of Sb4O6 produced will depend only on the moles of O2 present (it is the limiting reactant). So, let us find moles of O2 present:

1.87 g O2 x 1 mole O2/32 g O2 = 0.0584 moles O2

Now, we can use moles of O2 to find moles of Sb4O6 produced:

0.0584 moles O2 x 1 mole Sb4O6 / 3 moles O2 = 0.0195 moles Sb4O6 theoretically formed

Converting this to grams, we get the theoretical yield (NOT the actual yield as asked in the question):

0.0195 moles Sb4O6 x 292 g Sb4O6 / mole Sb4O6 = 5.69 g Sb4O6 Theoretical yield

(NOTE: you cannot calculate the actual yield as that is something that is measured in the lab)