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Stoichiometric Calculations

A 3.0 g sam...

Question

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.

95

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98

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92

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89

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Solution

The correct option is C $95$
$(C u^{+})_{2} ⟶ C u^{2 +} + C u$
The solution after dissolution of $C u_{2} O$ in $d i l . H_{2} S O_{4}$ contains $C u^{2 +}$ and $C u$ ions. $C u^{2 +}$ ions react with $K I$ to give $C u I_{2}$ , which is converted to $C u I$ and $I_{2}$ .
$C u^{2 +} + 2 I^{-} ⟶ C u I_{2} ⟶ C u I + \frac{1}{2} I_{2}$
MIllimole of $K I$ taken $= \frac{8.3}{166} \times 1000 = 50$
Now, $K I$ left unused reacts with oxidizing agent to liberate $I_{2}$ again.
$2 I^{-} O x i d a n t - ---- \to I_{2} 2 N a_{2} S_{4} O_{3} - ----- \to N a_{2} S_{4} O_{6} + 2 N a I$
$∴$ Millimole of $K I$ left $=$ Millimole of $N a_{2} S_{2} O_{3}$ used (mole ratio of $I^{-}$ to $N a_{2} S_{2} O_{3} i s 1 : 1)$ $= 10 \times 1.0 = 10$
Therefore, millimoles of $K I$ used for $C u_{2} O = 50 - 10 = 40$
$∴$ Millimoles of $C u_{2} O = 20$ $[C u_{2}^{+} ⟶ 2 C u^{2 +} ∴$ mole ratio $C u_{2}^{+} : C u^{2 +} :: 1 : 2]$
or, $\frac{w}{M} \times 1000 = 20$ $= \frac{w \times 1000}{142} = 20$
$∴ w_{C u_{2} O} = 2.84$
$∴ % o f C u_{2} O = \frac{2.84 \times 100}{3} = 94.67$

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Similar questions

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 :

0.32

g sample of impure

K I

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millimoles of

K_{2} C r O_{4}

and

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mL of

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H_{2} S O_{4}

. Iodine formed was expelled off by boiling and the solution is now mixed with excess of pure

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2 C r O_{4}^{2 -} + 2 H^{+} ⟶ C r_{2} O_{7}^{2 -} + H_{2} O

C r_{2} O_{7}^{2 -} + 6 I^{-} + 14 H^{+} ⟶ 2 C r^{3 +} + 3 I_{2} + 7 H_{2} O

2.5

g sample of copper is dissolved in excess of

H_{2} S O_{4}

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100

ml of

0.02 M C u S O_{4} (a q)

10

ml of

0.02 M

solution of

C u S O_{4} (a q)

is mixed with excess of

K I

to show the following changes:

C u S O_{4} + 2 K I ⟶ K_{2} S O_{4} + C u I_{2}

2 C u I_{2} ⟶ C u_{2} I_{2} + I_{2}

The liberated iodine is titrated with hypo (

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

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V

ml of

0.1 M

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I_{2}

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10

ml of

0.02 M

solution with excess

K I

is :

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

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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

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50

mL of

\frac{M}{10}

K I O_{3}

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20

mL of the same stock solution of

K I

requires

30

mL of

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

1.249

g of a sample of pure

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and impure

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was treated with dilute

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168

mL of

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10

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

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