Question

f a polymer is prepared from 5.54862 g of acrylic acid, and 2 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

Let's write the overall reaction:

r: C₂H₃COOH + NaOH ----------> C₂H₃COONa + H₂O

Let's calculate the moles of both reactants to determine the limitant reactant:

moles acid = 5.54862 / 72.06 = 0.077 moles

moles base = 8 * 0.002 = 0.016 moles

acid > base, therefore the moles of base are the limitant so:

r: C₂H₃COOH + NaOH ----------> C₂H₃COONa + H₂O

i: 0.077 0.016 0

e: 0.077-0.016 0 0.016

mass of C₂H₃COONa = 0.016 moles * 94.05 g/mol = 1.5048 g.

Hope this helps