Percent Yield and Limiting Stoichiometry. EXAMPLE QUESTION #1 (Limiting Stoichiometry ICE BOX)Limiting Stoichiometry Calculate the theoretical yield of.

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Percent Yield and Limiting Stoichiometry

EXAMPLE QUESTION #1 (Limiting Stoichiometry ICE BOX)Limiting Stoichiometry Calculate the theoretical yield of C 2 H 5 Cl if 112 g of C 2 H 5 OH is reacted with 34.7g of PCl 3 based on the reaction below. If 23.7 g of C 2 H 5 Cl is produced, what is the percent yield? –3 C 2 H 5 OH + PCl 3 ==> 3 C 2 H 5 Cl + H 3 PO 3

Step 1 – SET UP the ICE box 3 C 2 H 5 OH+ PCl 3 ==> 3 C 2 H 5 Cl+ H 3 PO 3 I-Initial C-Change-3X-X+3X+X E-End STEP 2- Find the moles, This is where you have to problem solve. 112g C 2 H 5 OH x (1 mole C 2 H 5 OH / 46.0g C 2 H 5 OH) = 2.43 moles C 2 H 5 OH 34.7g PCl 3 x (1mole PCl 3 / 137.5g PCl 3 ) = moles PCl 3

STEP 3- Find X, find the moles of everything If C 2 H 5 OH is limiting; X= 0; X= If PCl 3 is limiting X= 0; X= Therefore, PCl 3 is Limiting and X=0.252 (it is smaller) 3 C 2 H 5 OH+ PCl 3 ==> 3 C 2 H 5 Cl+ H 3 PO 3 I-Initial C-Change-3X-X+3X+X E-End 3 C 2 H 5 OH+ PCl 3 ==> 3 C 2 H 5 Cl+ H 3 PO 3 I-Initial C-Change -3(0.252)= (0.252)+(0.252) E-End

STEP 4- Answer the questions, convert moles to mass Theoretical mass of C 2 H 5 Cl produced is –0.756mol (64.5g/mol)= 48.8g C 2 H 5 Cl STEP 5- Percent Yield- Now take out that mass recovered and plug into the % yield equation. % yield = (actual yield/ theoretical yield)x 100 ‒ (23.7g/48.8) x 100%= 48.6% yield

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