What volume of $C O_{2}$ at NTP will be liberated by the action of $100 m L$ of $0.2 N H C l$ on $C a C O_{3}$ ?

112 m L

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224 m L

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448 m L

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120 m L

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Solution

The correct option is A $224 m L$
$R e a c t i o n w i l l l o o k l i k e t h i s,$
$C_{a} C O_{3} + 2 H C l \to C_{a} C l_{2} + C O_{2} + H_{2} O$
$1 m o l 2 m o l 1 m o l$

$N o r m a l i t y = \frac{g r a m e q .}{V o l . (L)} = \frac{g r a m e q .}{0.1}$
$(H C l)$
$0.1 \times 0.2 N = g r a m e q . \Rightarrow 0.02$
$M o l e s o f H C L = 0.02$
$M o l e s o f C_{a} C O_{3} = \frac{0.02}{2} = 0.01 m o l e$
$∴ 1 m o l o f C_{a} C O_{3} \to 22.4 L . o f C O_{2}$
$0.01 m o l o f C_{a} C O_{3} \to 22.4 \times 0.01$
$\Rightarrow 0.224 L . = 224 m L$

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