The reaction routinely gives a 75% yield of product. How much acetyl chloride an

Question

The reaction routinely gives a 75% yield of product. How much acetyl chloride and p-aminophenol
would you have to start with (in moles, grams and mL) if you wanted to end up with 20 g of the
product? Assuming a 75% yield amine salt, how much triethylamine hydrochloride (in moles and
grams) will be produced?
Densities acetyl chloride 1.10 g/mL; p-aminophenol 1.13 g/mL

PLEASE EXPLAIN

The reaction routinely gives a 75% yield of product. How much acetyl chloride and p-aminophenol would you have to start with (in moles, grams and mL) if you wanted to end up with 20 g of the product? Assuming a 75% yield amine salt, how much triethylamine hydrochloride (in moles and grams) will be produced? Densities - acetyl chloride 1.10 g/mL; p-aminophenol 1.13 g/mL

Explanation / Answer

the reaction gives

2 acetyl chloride + p-aminophenol   ---------->   amine salt + 2 triethylamine hydrochloride

Mol. weight of salt = 11 + 16 * 3 + 14 + 12 * 10 = 193 gm

20 gm = 20/193 mol = 0.1036 mol

As it gives 75% yield therfore the moles of p-aminophenol = 0.1036 mol * 100/75 = 0.1382 mol = 109.126 * 0.1382 mol = 15.077gm = 15.077gm/1.13 = 13.343 ml

moles of acetyl chloride= 0.1382 mol * 2 = 0.2764mol = 78.49 g/mol * 0.2764 = 21.7 = 21.7/1.10 g/mL = 19.72 ml

moles of triethylamine hydrochloride = 0.1036 mol *2 = 0.2072 mol = 136 * 0.2072 gm = 28.18 gm

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