Question

Calculate the percentage of available chlorine in a given sample of bleaching powder from the following data.
3.55g of bleaching powder when treated with acetic acid and excess of KI liberated iodine which required 60mL of 0.5N sodium thiosulphate solution.

Answer 1

Calculate the mass of bleaching powder that reacted out of 3.55g and then change it to mass of chlorine that reacted. Finally, divide it by the mass of bleaching powder to get the fraction of available chlorine.

Complete step by step answer:
Calculating the mass of bleaching powder that reacted.
Step 1) The balanced chemical equations for the given chemical reactions are,
$CaOC{l_2} + 2C{H_3}COOH + 2KI \to 2C{H_3}COOK + CaC{l_2} + {I_2} + {H_2}O$
$2N{a_2}{S_2}{O_3} + {I_2} \to N{a_2}{S_4}{O_6} + 2NaI$
From the coefficients of balanced chemical equations, it is clear that one (g-equiv) of bleaching powder ($CaOC{l_2}$) liberates one (g-equiv) of Iodine, which in turn requires $2$ (g-equiv) of sodium thiosulphate to completely react.
Therefore, Number of moles of $CaOC{l_2}$ = Number of moles of ${I_2}$= $\dfrac{1}{2}$ Numbers of moles of $\left( {N{a_2}{S_2}{O_3}} \right)$
Numbers of moles of $CaOC{l_2} = \dfrac{1}{2}\left( {\dfrac{{\left( {0.5g\; equiv } \right)}}{{1000mL}} \times 60mL} \right) \times \dfrac{{1{\text{ }}mole{\text{ }}of\;CaOC{l_2}}}{{1g{\text{ }}equiv.{\text{ }}of\;CaOC{l_2}}}$
Numbers of moles of $CaOC{l_2} = 0.015\;moles$
Now,
Numbers of moles of Cl in $CaOC{l_2} = 2 \times \left( {Numbers{\text{ }}of{\text{ }}moles{\text{ }}ofCaOC{l_2}} \right)$
Numbers of moles of $Cl = 2 \times 0.015 = 0.03\;moles$
Mass of 0.03 moles of Cl $= 35.5 \times 0.03 = 1.065\;g$
That is, 1.065g of Cl was available to react and produce ${I_2}$.
Step 2) Calculating the percentage of Chlorine available in a given mass of bleaching powder,
The percentage of chlorine available can be calculated by the formula:
Percentage of Chlorine available$= \dfrac{{Chlorine{\text{ }}that{\text{ }}reacted{\text{ }}}}{{Given{\text{ }}mass{\text{ }}of{\text{ }}bleaching{\text{ }}powder}} \times 100$
Both total chlorine and chlorine that reacted have been calculated in Steps 1 and 2 respectively.

Therefore, Percentage of Chlorine available $= \dfrac{{1.065g}}{{3.55g}} \times 100$ = 30.0 %

Note: Have an idea of molar ratios by writing balanced chemical equations and when the reacting moles of bleaching powder is found, do not forget to double it as each mole of bleaching powder contains two moles of chlorine. The g-equivalents of bleaching powder arte equal to the number of moles because it gives only one oxygen atom per molecule.