H+ Ferrocene + 2 acetyl chloride -------> diacetylferrocene 225 mg of ferrocene

Question

H+

Ferrocene + 2 acetyl chloride -------> diacetylferrocene

225 mg of ferrocene is followed to react with 0.140 mL of acetyl chloride in the presence of the catalyst, phosphoric acid, in hopes of synthesizing a target molecule called diacetylferrocene according to the balanced organic reaction shown below. After completing the reaction, the experimenters collect 164 mg of diacetylferrocene.

1. Find the limiting reactand and show all of your work

2. Calculate the theoretical yeild and percent yield of diacetylferrocene..

Explanation / Answer

1. ferrocene + 2 acetyl chloride => diacetylferrocene

Theoretical moles of ferrocene : acetyl chloride = 1 : 2


Moles of ferrocene = mass/molar mass of ferrocene

= 0.225/186.04 = 0.0012094 mol


Mass of acetyl chloride = volume x density

= 0.140 x 1.104 = 0.15456 g

Moles of acetyl chloride = mass/molar mass of acetyl chloride

= 0.15456/78.49 = 0.0019692 mol


Experimental moles of ferrocene : acetyl chloride

= 0.0012094 : 0.0019692 = 0.61415 : 1 = 1.2283 : 2


Since ferrocene is in excess

=> acetyl chloride is the limiting reactant

2. Moles of diacetylferrocene = 1/2 x moles of acetyl chloride

= 1/2 x 0.0019692 = 0.0009846 mol


Theoretical yield = moles x molar mass of diacetylferrocene

= 0.0009846 x 270.10

= 0.266 g = 266 mg


Percent yield = actual yield/theoretical yield x 100%

= 164/266 x 100%

= 61.7%

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