Finding the limiting reagent, theoretical yield, and percent yield. PLEASE EXPLA

Question

Finding the limiting reagent, theoretical yield, and percent yield.

PLEASE EXPLAIN! Thank you! Questions are below

In a similar experiment to the one being performed in the lab, 3.75 g of ammonium chloride (MM = 53.49 g/mol, D = 1.53 g/mL) is reacted with 2.42 mL of t-butanol (MM = 74.12 g/mol, D = 0.78 g/mL) to form t-butyl chloride (MM = 92.57 g/mol). After the reaction 1.60 g of t-butyl chloride is synthesized. Find the limiting reagent, theoretical yield and percent yield of t-butyl chloride. Note that in the actual experiment you will have two nucleophiles making it impossible to find theoretical yield. Limiting Reagent= Theoretical Yield = g Percentage Yield = %

Explanation / Answer

The reaction is

C4H9OH + NH4Cl --> C4H9Cl

As clear from the stoichiometry 1 mole of butanol will react with one mole of ammonium chloride to give one mole of butyl chloride.

Molecular weight of NH4Cl = 53.49 , amount used = 3.75 g so moles of ammonium chloride used = 0.07 moles

Molecular wt of butanol = 74.12 , amount used = 2.42 mL = 1.89 g; so moles of butanol used = 0.025 moles

Here the limiting reagent is one which is present in lesser amount , so t-butanol is the limiting reagent.

The two will react to give 0.025 moles of butyl chloride

So theoretical yield = moles X mol wt = 0.025 X 92.75 = 2.32 g

amount synthesized = 1.60

So % yield = actual yield X 100/ theoretical yield = 68.96%

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