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Percent Yield and Limiting Stoichiometry

 
 

 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

  3 C2H5OH +  PCl3 ==> 3 C2H5Cl  +   H3PO3
I-Initial    

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C-Change -3X -X +3X +X
E-End        

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

  3 C2H5OH +  PCl3 ==> 3 C2H5Cl  +   H3PO3
I-Initial 2.43 mol 0.252 mol

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C-Change -3X -X +3X +X
E-End        

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)

  3 C2H5OH +  PCl3 ==> 3 C2H5Cl  +   H3PO3
I-Initial 2.43 mol 0.252 mol

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C-Change -3(0.252) -(0.252) +3(0.252) +(0.252)
E-End 1.67mol 0 0.756mol 0.252mol

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

  3H2 +N2 ==> 2NH3
I-Initial    

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C-Change -3X -X +2X
E-End      

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

  3H2 +N2 ==> 2NH3
I-Initial

0.900

0.350

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C-Change -3X -X +2X
E-End      

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)

  3H2 +N2 ==> 2NH3
I-Initial

0.900

0.350

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C-Change -3(0.300) -(0.300) +2(0.300)
E-End 0 0.050 mol 0.600mol

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

 

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