# Easy Stoichiometry

Chemistry Level 4

$$20$$ ml of $$\dfrac{1}{60}\text{ M}$$ solution of $$\text{KBrO}_{3}$$ was added to a definite volume of $$\text{SeO}_{3}^{2-}$$ solution . The Bromine gas evolved was removed by boiling and excess of $$\text{KBrO}_{3}$$ was back titrated with $$5$$ mL of $$\dfrac{1}{25}\text{ M}$$ $$\text{NaAsO}_{2}$$ .

The Chemical Reactions taking place are as follows :

• $$\text{SeO}_{3}^{2-} + \text{BrO}_{3}^{-} + \text{H}^{+} \longrightarrow \text{SeO}_{4}^{2-} + \text{Br}_{2} + \text{H}_{2}\text{O}$$

• $$\text{BrO}_{3}^{-} + \text{AsO}_{2}^{-} + \text{H}_{2}\text{O} \longrightarrow \text{Br}^{-} + \text{AsO}_{4}^{3-} + \text{H}^{+}$$

Calculate the amount (in milligrams) of $$\text{SeO}_{3}^{2-}$$ in the solution

Some Important Data:

Atomic masses: K=39, Br=80, As=75, Na=23, O=16 and Se=79

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