EXAMPLE QUESTION #1 (Limiting Stoichiometry ICE BOX)

Calculate the theoretical yield of C₂H₅Cl if 112 g of C₂H₅OH is reacted with 34.7g of PCl₃ based on the reaction below.  If 23.7 g of C₂H₅Cl is produced, what is the percent yield?

 3 C₂H₅OH  +     PCl₃ ==> 3 C₂H₅Cl  +     H₃PO₃

_{(put 23.7g in your pocket for later)}

STEP1= SET UP the ICE Box

STEP 2- Find the moles, This is where you have to problem solve.

112g C₂H₅OH x (1 mole C₂H₅OH / 46.0g C₂H₅OH) = 2.43 moles C₂H₅OH

34.7g PCl₃ x (1mole PCl₃ / 137.5g PCl₃) = 0.252 moles PCl₃

STEP 3- Find X, find the moles of everything

If C₂H₅OH is limiting;   2.43 -3X= 0;  X= 0.810

If PCl₃ is limiting    0.252- X= 0;  X= 0.252

Therefore, PCl₃ is Limiting and X=0.252 (it is smaller)

STEP 4- Answer the questions, convert moles to mass

Theoretical mass of C₂H₅Cl produced is

0.756mol (64.5g/mol)= 48.8g C₂H₅Cl

STEP 5- Percent Yield- Now take out that mass recovered and plug into the % yield equation.

(23.7g/48.8) x 100%= 48.6% yield

EXAMPLE QUESTION #2 (Limiting Stoichiometry ICE BOX)

1.80 g H₂ is allowed to react with 9.79 g N₂, producing 2.02 g NH₃.

What is the theoretical yield and the percent yield for this reaction under the given conditions?

THE BALANCED REACTION   3H₂ + N₂ ==> 2NH₃

See how it says yields? Actually ==>producing 2.02 g NH₃ save that for later (in your pocket).

STEP1= SET UP the ICE Box

STEP 2- Find the moles, This is where you have to problem solve.

1.80g H₂ x (1 mole H₂ / 2.0g H₂) = 0.900 moles H₂

9.79g N₂ x (1mole N₂ / 28.02g N₂) = 0.350 moles N₂

STEP 3- Find X, find the moles of everything

If H₂ is limiting;   0.900 -3X= 0;  X= 0.300

If N₂ is limiting    0.350- X= 0;  X= 0.350

Therefore  H₂ is Limiting and X=0.300 (it is smaller)

STEP 4- Answer the questions, convert moles to mass

Theoretical mass of NH₃ produced is 0.600mol (17.0g/mol)= 10.2g

STEP 5- Percent Yield- Now take out that mass recovered and plug into the % yield equation.

(2.02g/10.2g) x 100%= 19.8% yield