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JEE Main levelSingle correct optionDifficulty 2 of 3ph097

$17.4 g$ of $MnO X_{2}$ ( $M = 87 g mo l^{- 1}$ ) is heated with excess concentrated $HCl$ . Taking the molar volume at STP as $22.4 L mo l^{- 1}$ , the volume of $Cl X_{2}$ evolved at STP is

A $1.12 L$
B $4.48 L$
C $2.24 L$
D $8.96 L$

Answer: B

MnO X_{2} + 4 HCl MnCl X_{2} + Cl X_{2} + 2 H X_{2} O

gives 1 mol

Cl X_{2}

per mol

MnO X_{2}

. Moles of

MnO X_{2} = 17.4/87 = 0.2

, so

V = 0.2 \times 22.4 = 4.48 L

.

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