Question

* Densities of the two reactants are: Sebacoyl chloride 1.121 g/mL and, 1,6-diaminohexane 0.83 g/mL

yion Co1o o-diaminohexane Se ba coyl chloride If I start with 2 mL of the sebacoyl chloride and 1.3 mL 1,6- diaminohexane and end up with 3.30g of my product nylon 6,10 what is my percent yield?

Answer 1

We know

density = mass/ volume

mass = density * volume

Mass of Sebacoyl chloride = 1.121 g/mL * 2ml = 2.242g

Mass of 1,6-diaminohexane = 0.83 g/mL* 1.3ml = 1.079g

molar mass of 1,6-diaminohexane =116.21 g/mol

molar mass of Sebacoyl chloride = 239.14g/mol

molar mass of nylon 6,10 = 318.458 g/mol

According to reaction;

239.14g of Sebacoyl chloride reacts with 116.21 g of 1,6-diaminohexane

so, 2.242g  of Sebacoyl chloride reacts with = (116.21/239.14)2.242 g of 1,6-diaminohexane

= 1.09g of 1,6-diaminohexane

as only 1.079g of 1,6-diaminohexane is taken, hence 1,6-diaminohexane is limiting reagent.

now , according to reaction ; Theoritical yield  of nylon 6,10 is:-

116.21 g of 1,6-diaminohexane gives  = 318.458 g of nylon 6,10

so, 1.079g of 1,6-diaminohexane will give = (318.458 g /116.21 g) * 1.079 g  of nylon 6,10

= 2.96g  of nylon 6,10

percent yield = (experimental yield/Theoritical yield ) *100

= (3.30/2.96) * 100

= 111.5%

but it is practically not possible,

practical yield can never be more than that of theritical yield.

hope this is helpful!