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Stoichiometric Calculations

One gram of c...

Question

One gram of commercial $A g N O_{3}$ is dissolved in 50 mL of water. It is treated with 50 mL of a KI solution. The silver iodide thus precipitated in filtered off. Excess of KI in the filtrate is titrated with $(M / 10) K I O_{3}$ solution in presence of 6 M HCl till all $I^{-}$ ions are converted into ICI. It requires 50 ml. of $(M / 10) K I O_{3}$ solution. 20 mL of the same stock solution of KI requires 30 mL of $(M / 10) K I O_{3}$ under similar conditions. Calculate the percentage of $A g N O_{3}$ in the sample.
(Reaction: $K I O_{3} + 2 K I + 6 H C l \to 3 I C I + 3 K C l + 3 H_{2} O$ )

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Solution

Reaction involved titration is $K I O_{3} + 2 K I + 6 H C l \to 3 I C l + 3 K C l + 3 H_{2} O$
1 mole 2 mole

$mmoles of KI solution in original 20 mL of stock solution = 2 \times mmoles of K I O_{3} in 30 mL solution$

$total mmoles of KI in orignal solution = 2 \times 0.1 \times \frac{30 \times 50}{20}$

Millimoles in 50 ml. of KI solution $= 15$

Millimoles of KI left unreacted with $A g N O_{3}$ solution $= 2 \times 50 \times \frac{1}{10} = 10$

$∴$ Millimoles of KI reacted with $A g N O_{3}$
Millimoles of $A g N O_{3}$ present $A g N O_{3}$ solution = 5
Molecular weight of $A g N O_{3}$
$∴$ Wt. of $A g N O_{3}$ in the solution $= 5 \times 10^{- 3} \times 170 = 0.850$ g
% $A g N O_{3}$ in the sample $= \frac{0.850}{1} \times 100 = 85$ %

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Similar questions

Q. One gram of commercial

A g N O_{3}

is dissolved in 50 mL of water. It is treated with 50 mL of a KI solution. The silver iodide thus precipitated in filtered off. Excess of KI in the filtrate is titrated with

(M / 10) K I O_{3}

solution in presence of 6 M HCl till all

I^{-}

ions are converted into ICI. It requires 50 ml. of

(M / 10) K I O_{3}

solution. 20 mL of the same stock solution of KI requires 30 mL of

(M / 10) K I O_{3}

A g N O_{3}

in the sample.
(Reaction:

K I O_{3} + 2 K I + 6 H C l \to 3 I C l + 3 K C l + 3 H_{2} O

)

1

g sample of

A g N O_{3}

is dissolved in

50

mL of water. It is titrated with

50

mL of

K I

solution. The

A g I

precipitated is filtered off. Excess of

K l

in filtrate is titrated with

\frac{M}{10}

K I O_{3}

in presence of

6 M H C l

till all

I^{-}

converted into

I C l

. It requires

50

mL of

\frac{M}{10}

K I O_{3}

solution.

20

mL of the same stock solution of

K I

requires

30

mL of

\frac{M}{10}

K I O_{3}

A g N O_{3}

in sample. The reaction is given below:

K I O_{3} + 2 K I + 6 H C l ⟶ 3 I C l + 3 K C l + 3 H_{2} O

Q. A

3.0 g

sample containing

F e_{3} O_{4}, F e_{2} O_{3}

and an inert impure substance is treated with excess of

K I

solution in presence of dilute

H_{2} S O_{4}

. The entire iron is converted to

F e^{2 +}

along with the liberation of iodine. The resulting solution is diluted to

100 m L

. A

20 m L

of the dilute solution requires

11.0 m L

0.5 M N a_{2} S_{2} O_{3}

solution to reduce the iodine present. A

50 m L

of the diluted solution after complete extraction of iodine requires

12.8 m L

0.25 M K M n O_{4}

solution in dilute

H_{2} S O_{4}

medium for oxidation of

F e^{2 +}

. Calculate the percentage of

F e_{2} O_{3}

and

F e_{3} O_{4}

in the original sample.

10.0

g sample of

C u_{2} O

is dissolved in

d i l . H_{2} S O_{4}

where it undergoes disproportionation quantitatively. The solution is filtered off and

8.3

g pure

K I

crystals are added to clear filtrate in order to precipitate

C u I

with the liberation of

I_{2}

. The solution is again filtered and boiled till all the

I_{2}

is removed. Now excess of an oxidising agent is added to the filtrate which liberates

I_{2}

again. The liberated

I_{2}

now requires

10

mL of

1.0 M

N a_{2} S_{2} O_{3}

solution. The percentage by mass of

C u_{2} O

in the sample is :

Q. A

3.0

g sample of

C u_{2} O

is dissolved in

d i l . H_{2} S O_{4}

where it undergoes disproportionation quantitatively. The solution is filtered off and

8.3

g pure

K I

crystals are added to clear filtrate in order to precipitate

C u I

with evolution of

I_{2}

. The solution is again filtered and boiled till all the

I_{2}

is expelled. Now excess of an oxidizing agent is added to filtrate, which liberates

I_{2}

again. The liberated

I_{2}

this time requires

10

mL of

1.0 N

N a_{2} S_{2} O_{3}

solution. Calculate

%

by mass of

C u_{2} O

in sample.

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