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Stoichiometric Calculations

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Question

Calculate the millimoles of $S e O_{3}^{2 -}$ in solution on the basis of following data:
$70 m l$ of $\frac{M}{60}$ solution of $K B r O_{3}$ was added to $S e O_{3}^{2 -}$ solution. The bromine evolved was removed by boiling and excess of $K B r O_{3}$ was back titrated with $12.5 m L$ of $\frac{M}{25}$ solution of $N a A s O_{2}$ .
The reactions are given below.
I. $S e O_{3}^{2 -} + B r O_{3}^{-} + H^{+} \to S e O_{4}^{2 -} + B r_{2} + H_{2} O$
II. $B r O_{3}^{-} + A s O_{2}^{-} + H_{2} O \to B r^{-} + A s O_{4}^{3 -} + H^{+}$

A

1.6 \times 10^{- 3}

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B

1.25

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C

2.5 \times 10^{- 3}

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D

None of these

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Solution

The correct option is C $2.5 \times 10^{- 3}$
(I) $S e + 4 O_{3}^{2 -} + + 5 B r O_{3}^{-} + H^{+} \to + 6 S e O_{4}^{2 -} + 0 B r_{2} + H_{2} O$
(II) $+ 5 B r O_{3}^{-} + + 3 A s O_{2}^{-} + H_{2} O \to - 1 B r^{-} + + 5 A s O_{4}^{3 -} + H^{+}$
In reaction $(I I)$ ,
gm. eq. of $B r O_{3}^{-} =$ gm. eq. of $A s O_{2}^{-}$
$n_{B r O_{3}^{-}} \times 6 = n_{A s O_{2}^{-}} \times 2 = \frac{12.5}{1000} \times \frac{1}{25} \times 2 = 10^{- 3}$

$n_{B r O_{3}^{-}} = \frac{10^{- 3}}{6}$

In reaction (I),
moles of $B r O_{3}^{-}$ consumed $= \frac{70}{1000} \times \frac{1}{60} - \frac{10^{- 3}}{6} = 10^{- 3}$

gm eq. of $S e O_{3}^{2 -} = g m$ . eq. of $B r O_{3}^{-}$
$n_{S e O_{3}^{2 -}} \times 2 = 10^{- 3} \times 5$
$n_{S e O_{3}^{2 -}} = 2.5 \times 10^{- 3}$

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Similar questions

Q. Calculate the amount (in milligrams) of

S e O_{3}^{2 -}

in solution on the basis of following data:

20

\frac{M}{60}

K B r O_{3}

was added to a definite volume of

S e O_{3}^{2 -}

solution. The bromine evolved was removed by boiling and excess of

K B r O_{3}

was back titrated with

5

\frac{M}{25}

N a A s O_{2}

. The reactions are given below.

1) S e O {_{3}}^{2 -} + B r O {_{3}}^{-} + H^{-} \to S e O {_{4}}^{2 -} + B r_{2} + H_{2} O

2) B r O {_{3}}^{-} + A s O_{2} + H_{2} O \to B r^{-} + A s O {_{4}}^{3 -} + H^{+}

(Atomic mass of

K = 39, B r = 80, A s = 75, N a = 23, O = 16, S e = 79

)

70 m l, M / 60 s o l^{n}

K B r O_{3}

S e O_{3}^{2 -}

. Bromine evolved in removed by boiling & excess of

K B r O_{3}

in back titrated with

12.5 m l M / 25 s o l^{n}

N a A s O_{2}

. Calculate millilitre of

S e O_{3}^{2 -}

Q. Calculate the mass of

S e O_{3}^{2 -}

in solution on the basis of following data.

20

\frac{M}{60}

K B r O_{3}

was added to a definite volume of

S e O_{3}^{2 -}

solution. The bromine evolved was removed by boiling and excess of

K B r O_{3}

was back titrated with

5.1

\frac{M}{25}

N a A s O_{2}

. The reactions are given below:

S e O_{3}^{2 -} + B r O_{3}^{-} + H^{+} ⟶ S e O_{4}^{2 -} + B r_{2} + H_{2} O

B r O_{3}^{-} + A s O_{2}^{-} + H_{2} O ⟶ B r^{-} + A s O_{4}^{3 -} + H^{+}

B r O_{3}^{-}

shows two reactions :

S e O_{3}^{2 -} + B r O_{3}^{-} + H^{+} ⟶ S e O_{4}^{2 -} + B r_{2} + H_{2} O

A s O_{2}^{2 -} + B r O_{3}^{-} + H^{+} ⟶ B r^{-} + A s O_{4}^{-} + H_{2} O

Reason: The ratio of equivalent masses of

B r O_{3}^{-}

in two reactions is

\frac{5}{6}

.

B r O_{3}^{-}

ion reacts with

B r^{-}

ion in acid solution and

B r_{2}

is liberated. The equivalent weight of

K B r O_{3}

in this reaction is:

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