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Standard XII

Chemistry

Stoichiometric Calculations

10 g of MnO...

Question

$10$ g of $M n O_{2}$ on reaction with conc. $H C l$ liberated $0.1$ equivalent of $C l_{2}$ (molecular weight of $M n = 55$ g/mol). Hence, percentage purity of $M n O_{2}$ is :

87.0

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21.75

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50.0

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43.5

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Solution

The correct option is D $43.5$
The balanced chemical reaction is as shown below.
0.1 equivalent of chlorine corresponds to 0.05 moles of chlorine.
10 g sample contains 0.05 mole of $M n O_{2}$ or $0.05 \times 86.9 \times 4.35 g M n O_{2}$
The percentage purity of the sample $= \frac{4.35}{10} \times 100 = 43.5$

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Similar questions

Q. A

0.5

g sample containing

M n O_{2}

on treatment with

H C l

liberates

C l_{2}

. The

C l_{2}

is passed into a solution of

K I

and

30.0

mL of

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

are required to titrate the liberated iodine. Calculate the percentage of

M n O_{2}

in the sample (as nearest integer).

[Given that the atomic weight of

M n = 55

g/mol]

Q. For the reaction :

M n O_{2} + 4 H C l 100 % - -- \to M n C l_{2} + C l_{2} + 2 H_{2} O

17.4 g M n O_{2}

is dissolved in

500 m L

H C l

solution containing

7.3 g

H C l

per litre. (molar mass of

M n = 55 g / m o l

)

0.5

g sample containing

M n O_{2}

is treated with

H C l

liberating

C l_{2}

. The

C l_{2}

is passed into a solution of

K I

and

30.0

cc of

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

are required to titrate the liberated iodine. Calculate the percentage of

M n O_{2}

in sample (atomic mass of

M n = 55

Q. Assertion :One equivalent of

M n O_{2}

reacts with 2 equivalents of

H C l

in the reaction,

M n O_{2} + 4 H C l ⟶ M n C l_{2} + 2 H_{2} O + C l_{2}

. Reason: One equivalent of

M n O_{2}

reacts with one equivalent of

H C l

Q. For the reaction :

M n O_{2} + 4 H C l 100 % - -- \to M n C l_{2} + C l_{2} + 2 H_{2} O

17.4 g M n O_{2}

is dissolved in

500 m L

H C l

solution containing

7.3 g

H C l

per litre. (molar mass of

M n = 55 g / m o l

)

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10 g of MnO2 on reaction with conc. HCl liberated 0.1 equivalent of Cl2 (molecular weight of Mn=55 g/mol). Hence, percentage purity of MnO2 is :

The correct option is D 43.5The balanced chemical reaction is as shown below.0.1 equivalent of chlorine corresponds to 0.05 moles of chlorine.10 g sample contains 0.05 mole of MnO2 or 0.05×86.9×4.35 g MnO2The percentage purity of the sample =4.3510×100=43.5

$10$ g of $M n O_{2}$ on reaction with conc. $H C l$ liberated $0.1$ equivalent of $C l_{2}$ (molecular weight of $M n = 55$ g/mol). Hence, percentage purity of $M n O_{2}$ is :