A student followed the procedure of this experiment to determine the percent NaO

Question

A student followed the procedure of this experiment to determine the percent NaOCl in a commercial bleaching solution that was found in the basement of an abandoned house. The student diluted 50.00 mL of commercial bleaching solution to 250 mL in a volumetric flask, and titrated a 20-mL aliquot of the diluted bleaching solution. The titration required 35.46 mL of 0.1052M Na2S2O3 solution. A faded price label on the gallon bottle read $0.79. The density of the bleaching solution was 1.10 g/mL

Calculate the number of moles of I2 produced in the titration mixture.

Explanation / Answer

Molarity of Na2S2O3 = 0.1052 M

Volume = 35.46 mL

Density of bleaching solution = 1.10 g /mL

Lets calculate moles Na2S2O3

Moles of Na2S2O3 = volume of Na2S2O2 in L * molarity

=0.03546 L * 0.1052 M

= 0.00373 mol

Reaction :

We calculate moles of NaOCl in the reaction mixture

4 NaOCl + Na2S2O3 + 2 NaOH --- > 4 NaCl + 2 Na2SO4 + H2O

Here from the reaction stoichiometry we say that

1 mol of sodium thiosulfate needs 4 mol NaOCl

Mol of NaOCl required to react with 0.00375 mol thiosulfate

= 0.00375 mol thiosulfate * 4 mol NaOCl / 1 mol thiosulfate

= 0.0149 mol NaOCl

To find moles of I2 we write the reaction

NaOCl + 2I - + 2 H3O --- > I2+ NaCl + H2O

One mole NaOCl produces 1 mol I2

Mol of I2 produced from 0.0149 mol NaOCl

= 0.0149 mol NaOCl * 1 mol I2 / 1 mol NaOCl

= 0.0149 mol I2

Moles of I2 produced = 0.0149 mol

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