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What is the theoretical yield of nylon-6,6 that can be produced from the reaction of 25 grams of adipic acid and 35 grams of hexamethylenediamine? (Assuming 100% reaction efficiency)

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To calculate the theoretical yield of nylon-6,6, we first need to determine the limiting reactant in the reaction between adipic acid and hexamethylenediamine. We will do this by comparing the moles of each reactant.1. Calculate the molar mass of adipic acid  C6H10O4  and hexamethylenediamine  C6H16N2 :Adipic acid:  6 x 12.01  +  10 x 1.01  +  4 x 16.00  = 146.14 g/molHexamethylenediamine:  6 x 12.01  +  16 x 1.01  +  2 x 14.01  = 116.21 g/mol2. Calculate the moles of each reactant:moles of adipic acid = 25 g / 146.14 g/mol = 0.171 molesmoles of hexamethylenediamine = 35 g / 116.21 g/mol = 0.301 moles3. Determine the limiting reactant:The reaction between adipic acid and hexamethylenediamine occurs in a 1:1 ratio. Since there are fewer moles of adipic acid, it is the limiting reactant.4. Calculate the theoretical yield of nylon-6,6:The molar mass of nylon-6,6  C12H22N2O2  is  12 x 12.01  +  22 x 1.01  +  2 x 14.01  +  2 x 16.00  = 226.32 g/mol. Since the reaction occurs in a 1:1 ratio, the moles of nylon-6,6 produced will be equal to the moles of the limiting reactant  adipic acid .moles of nylon-6,6 = 0.171 molestheoretical yield of nylon-6,6 = 0.171 moles x 226.32 g/mol = 38.7 g  rounded to one decimal place So, the theoretical yield of nylon-6,6 that can be produced from the reaction of 25 grams of adipic acid and 35 grams of hexamethylenediamine is 38.7 grams, assuming 100% reaction efficiency.
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