Question

50 g caustic soda is completely converted into sodium chlorate and sodium chloride by the action of chlorine. If X mL volume of HCl (containing 300g \[{{\text{L}}^{-\text{1}}}\]) were used for the production of necessary chlorine, then the value of 100X is:
a.) 303.33
b.) 3300
c.) 3350
d.) None of these

Answer 1

Hint: We should first write the reaction of formation of chlorine. And then we will react to the chlorine with caustic soda.

Complete step by step solution:
We will first write the reaction for the production of chlorine.

\[\text{Mn}{{\text{O}}_{\text{2}}}+\text{4HCl}\to \text{MnC}{{\text{l}}_{\text{2}}}+\text{C}{{\text{l}}_{\text{2}}}+\text{2}{{\text{H}}_{\text{2}}}\text{O}\]

Now, we will write the molecular mass.
\[\text{Mn}{{\text{O}}_{\text{2}}}\]= 87 gram
\[\text{4HCl}\]= 4\[\times \]36.5 gram
\[C{{l}_{2}}\]= 71 gram
\[\text{3C}{{\text{l}}_{\text{2}}}+\text{6NaOH}\to \text{5NaCl}+\text{NaCl}{{\text{O}}_{\text{3}}}+\text{3}{{\text{H}}_{\text{2}}}\text{O}\]
\[C{{l}_{2}}\]=3\[\times \]71 gram
NaOH= 6\[\times \]40
We should know that, 50g of NaOH, the weight of \[C{{l}_{2}}\]required =\[\dfrac{3\times 71\times 50}{6\times 40}=44.25\,g\]
Weight of HCl required=\[\dfrac{4\times 36.5\times 44.25}{71}=91\,g\]
Now, we will find weight of \[\text{Mn}{{\text{O}}_{\text{2}}}\]required= \[\dfrac{87\times 91}{36.5\times 4}=54.226\,g\]
Now, we will find volume of HCl required= \[\dfrac{1000\times 91}{300}=303.33\,mL\]
So, from the discussion above we can say that option A is correct.


Note: We should know that mixture of almost any solid chloride and manganese dioxide (\[\text{Mn}{{\text{O}}_{\text{2}}}\]) yields chlorine when heated with concentrated sulphuric acid. We should note that most of the chlorine produced is used for chemical processes involving the introduction of chlorine into organic compounds, yielding carbon tetrachloride.