For the following reaction, 5.78 grams of chlorine gas are mixed with excess iron. The reaction yields 6.79 grams of iron(III) chloride. 2Fe (s) + 3Cl2 (g) → 2FeCl3 (s)

a. What is the theoretical yield of iron(III) chloride?
b. What is the percent yield for this reaction?

a. Theoretical yield = 8.81 g

b. 77 %

Explanation:

We begin from the reaction:

2Fe (s) + 3Cl₂ (g) → 2FeCl₃ (s)

and the excess is the iron, so the limiting reagent is the chlorine.

We convert mass to moles → 5.78 g / 70.90 g/mol = 0.0815 mol

Ratio is 3:2 so we think this:

From 3 moles of chlorine, I can produce 2 moles of Iron (III) chloride

Then 0.0815 moles would produce (0.0815 . 2) /3 = 0.0543 moles

This is the theoretical yield. We convert the moles to mass:

0.0543 mol . 162.2 g /mol = 8.81 g

If the reaction yields 6.79 g of FeCl₃ and the theoretical yield is 8.81 g ,we can obtain the percent yield: (6.79 / 8.81) . 100 = 77 %