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Standard XII

Chemistry

Limiting Reagent or Reactant

8.7 g of pure...

Question

8.7 g of pure $M n O_{2}$ is heated with an excess of HCI and the gas evolved is passed into a solution of KI. Calculate the amount of the iodine liberated (Mn = 55, Cl = 35.5, I =127):

0.1 mol

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25.4 g

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15.4 g

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7.7 g

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Solution

The correct option is A 25.4 g
we have,
$M n O_{2} + H C l \to C l_{2} + M n C l_{2} + H_{2} O$
and $2 K I + C l_{2} \to K C l + I_{2}$
molecular weight of $M n O_{2}$ = 87 g
no. of moles of $M n O_{2}$ = 0.1 mole
so, from ist equation we have 1 mole of $M n O_{2}$ produce 1 mole of $C l_{2}$ and from 2nd equation 1 mole of $C l_{2}$ is producing;
so, 0.1 mole of $M n O_{2}$ produces 0.1 mole of $I_{2}$
amount of $I_{2}$ produces = moles * molecular mass = 0.1 * 254 = 25.4 g

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Similar questions

8.7 g

of pure

M n O_{2}

is heated with an excess of

H C l

and the gas evolved is passed into a solution of

K I

. Calculate the weight of the iodine librated

(M n = 55, C l = 35.5. I = 127)

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g of pyrolusite (impure

M n O_{2}

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8.7 g of pure MnO2 is heated with an excess of HCI and the gas evolved is passed into a solution of KI. Calculate the amount of the iodine liberated (Mn = 55, Cl = 35.5, I =127):

8.7 g of pure $M n O_{2}$ is heated with an excess of HCI and the gas evolved is passed into a solution of KI. Calculate the amount of the iodine liberated (Mn = 55, Cl = 35.5, I =127):