How many grams with a percent yield

Question

How many grams of Fe(OH)2 were isolated from the reaction of 85.0 mL of a 0.360 M NaOH solution with excess Fe(ClO3)2, if the reaction occurred with an 86.9% yield?

Answer 1

First step: Write a balanced equation

 

2NaOH + Fe(ClO₃)₂ --> Fe(OH)_2 (s) + 2NaClO₃

 

Second step: determine your limiting reagent, which in this case is NaOH since Fe(ClO3)2 is said to be in excess

 

Third step: Calculate the theoretical moles of Fe(OH)2 

 

0.36 mol/L NaOH × 0.085 L = 0.0306 mol NaOH

0.0306 mol NaOH × 1 mol Fe(OH)₂ / 2 mol NaOH = 0.0153 mol Fe(OH)₂    

 

Fourth step: Convert to grams of Fe(OH)₂

 

0.0153 mol Fe(OH)₂ × 90 g / 1 mol Fe(OH)₂ = 1.377 g Fe(OH)₂

 

Fifth step: Find the actual yield

 

Percent yield = Actual / Theoretical × 100

 

therefore our actual yield would be Percent yield × theoretical 

which is 86.9/100 × 1.377 = 1.197