Question

# If a polymer is prepared from 4.10742 g of acrylic acid, and 3.625 mL of 8.0...

If a polymer is prepared from 4.10742 g of acrylic acid, and 3.625 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.

The polymer formed here is sodium acrylate.(C2H3COONa)

The moles of acrylic acid reacted =4.10742/72 =0.057 moles

The moles of NaOH reacted =volume * conc =0.003625 * 8 =0.029 moles
Acrylic acid + Sodium hydroxide ->forms-> Sodium acrylate + water
: C2H3COOH + NaOH -> C2H3COONa + H2O

1 mole NaOH reacts with 1 mole acrylic acid.

Here NaOH is limiting reactant.

So, moles of Sodium acrylate formed = moles of NaOH =0.029 moles

Mass of Sodium acrylate =0.029 moles * molar mass

=0.029 moles *94

=2.726 grams

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