In a similar experiment to the one being performed in the lab, 1.44 g of ammoniu
ID: 795286 • Letter: I
Question
In a similar experiment to the one being performed in the lab, 1.44 g of ammonium chloride (MM = 53.49 g/mol, D = 1.53 g/mL) is reacted with 3.81 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 2.15 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 = <--Make sure you have the correct spelling (including spaces and dashes as necessary). All compound names are spelled out in the question above.
Theoretical Yield = g
Percentage Yield = %
Explanation / Answer
ammonium chloride + t-butanol => t-butyl chloride + ammonium hydroxide
Theoretical moles of ammonium chloride : t-butanol = 1 : 1
Moles of ammonium chloride = mass/molar mass = 1.44/53.49 = 0.026921 mol
Moles of t-butanol = volume x density/molar mass = 3.81 x 0.78/74.12 = 0.040094 mol
Experimental moles of ammonium chloride : t-butanol = 0.026921 : 0.040094 = 1 : 1.4893
Since t-butanol is in excess => ammonium chloride is the limiting reagent
Moles of t-butyl chloride = moles of ammonium chloride = 0.026921 mol
Theoretical yield = moles x molar mass of t-butyl chloride
= 0.026921 x 92.57
= 2.492 g = 2.49 g
Percentage Yield = actual yield/theoretical yield x 100%
= 2.15/2.492 x 100%
= 86.3%
Thus:
Limiting Reagent = ammonium chloride
Theoretical Yield = 2.49 g
Percentage Yield = 86.3%