Question

If a polymer is prepared from 6.41334 g of acrylic acid, and 1.875 mL of 8.0 M NaOH, what will the mass of the "dry" polymer be after all of the water is removed? Assume all reactions go to completion. Hint: You may want to construct a BCA table to determine the outcome of the neutralization step.

Answer 1

1 mole of acrylic acid reacts with 1 mole of NaOH to produce 1 mole of water

thus, moles of acrylic acid reacting = mass/molar mass = 6.41334/72.06 = 0.089

moles of NaOH reacting = molarity*volume of solution in litres = 8*0.001875 = 0.015

mass of NaOH reacting = moles*molar mass = 0.015*40 = 0.6 g

Clearly, NaOH is limiting reagent

Thus, moles of acrylic acid unreacted = 0.089-0.015 = 0.074

mass of acrylic acid unreacted = moles*molar mass = 0.074*72.06 = 5.332 g

Thus, moles of polymer formed = 0.015 = moles of water formed

mass of water formed = moles*molar mass = 0.015*18 = 0.27 g

Thus, mass of polymer formed = total mass reacting - mass of unreacted acrylic acid - mass of water = 6.41334 + 0.6 - 5.332 - 0.27 = 1.41134 g