Nylone synthesis 6,10 We used 10 mL of a 1,6-hexanediamine/NaOH aqueous solution

Question

Nylone synthesis 6,10

We used 10 mL of a 1,6-hexanediamine/NaOH aqueous solution (0.5 M 1,6-hexanediamine/0.5 M NaOH) and pouring a 10 mL of a 0.2 M solution of decanedioyl chloride in hexane. I am lost how to do the calculation. For the theoretical yield the teacher said we need to add (in grams) the limiting reagent & the second reagent. I need detailed explanation for each calculation and these are the numbers I got from doing the experiment.

Diameter of beaker used for wrapping the nylon:0.055m

circumference of the beaker (c = d): 0.1727 m

number of turns of nylon around the beaker: 32

turns length of nylon rope: 5.5m

weight of nylon rope:1.8129 g

limiting reagent in reaction mixture: 0.2 M solution of decanedioyl chloride in hexane

weight of limiting reagent in reaction mixture in ?g

moles of limiting reagent in reaction mixture : ? moles

moles of second reagent in reaction mixture that should be consumed: ?moles

weight of second reagent in reaction mixture that should be consumed: ?g

theoretical yield of nylon: g

actual yield of nylon:g

% yield of nylon rope: g

Explanation / Answer

Molarity = no. of moles/volume (L)

The no. of moles of limiting reagent, i.e. decanedioyl chloride = 0.2 mol/L * 10*10-3 L = 2*10-3 mol

The no. of moles = mass/molar mass

The molar mass of decanedioyl chloride = 239 g/mol

i.e. The mass of decanedioyl chloride = 2*10-3 mol * 239 g/mol = 0.478 g

The no. of moles of second reagent, i.e.1,6-hexanediamine = 0.5 mol/L * 10*10-3 L = 5*10-3 mol

The molar mass of 1,6-hexanediamine = 116 g/mol

Therefore, the mass of 1,6-hexanediamine = 5*10-3 mol * 116 g/mol = 0.580 g

The theoretical yiled = 0.478 g + 0.580 g = 1.058 g

The given mass of nylon rope = 1.8129 g

i.e. The mass of 5.5 m/0.055 m = 100 turns of nylon rope = 1.8129 g

But the no. of turns of nylon rope around the beaker = 32

i.e. The actual yield of nylon obtained = (32/100)*1.8129 g = 0.581 g

The % yield of nylon rope = (actual yield/theoretical yield)*100

= (0.581 g/1.058 g)*100

= 54.9%

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