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Standard XII

Chemistry

Stoichiometric Calculations

Calculate the...

Question

Calculate the volume of $C l_{2} (g)$ (in mL) liberated at $1$ atm, $273$ K when $1.74$ gm $M n O_{2}$ reacts with $2.19$ gm $H C l$ according to the following reaction with $40 %$ yield.
$M n O_{2} + H C l \to M n C l_{2} + C l_{2} + H_{2} O$
(Atomic mass of $M n = 55, C l = 35.5$ )

336

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112

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134.4

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44.8

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Solution

The correct option is C $134.4$ mL
$M n O_{2} + 4 H C l \to M n C l_{2} + C l_{2} + 2 H_{2} O$
$\frac{1.74}{87}$ $\frac{2.19}{36.5}$
$= 2$ $= .06$
$0.06$ mol $H C l$ produces $\frac{1}{4} \times 0.06$ mol of $C l_{2}$
$= \frac{1}{4} \times .06 \times .4 = .006$ mol (with 40% yield).
Volume of $C l_{2} = \frac{n R T}{P} = 0.006 \times 22400$ $= 134.4$ mL.

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Calculate the volume of Cl2(g) (in mL) liberated at 1 atm, 273 K when 1.74 gm MnO2 reacts with 2.19 gm HCl according to the following reaction with 40% yield.MnO2+HCl→MnCl2+Cl2+H2O(Atomic mass of Mn=55,Cl=35.5)

The correct option is C 134.4 mLMnO2+4HCl→MnCl2+Cl2+2H2O1.7487 2.1936.5=2 =.060.06 mol HCl produces 14×0.06 mol of Cl2=14×.06×.4=.006 mol (with 40% yield).Volume of Cl2=nRTP=0.006×22400 =134.4 mL.

Calculate the volume of $C l_{2} (g)$ (in mL) liberated at $1$ atm, $273$ K when $1.74$ gm $M n O_{2}$ reacts with $2.19$ gm $H C l$ according to the following reaction with $40 %$ yield.
$M n O_{2} + H C l \to M n C l_{2} + C l_{2} + H_{2} O$
(Atomic mass of $M n = 55, C l = 35.5$ )