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100 g CaCO3...

Question

100 g $C a C O_{3}$ reacts with 1 litre 1M HCl. On completion of reaction how much weight of $C O_{2}$ will be obtain ?

5.5 g

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11 g

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22 g

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33 g

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Solution

The correct option is B 22 g
$C a C O_{3} + 2 H C l ⟶ C a {C l}_{2} + {C O}_{2} + H_{2} O$
$C a C O_{3}$ , Given Mass= $100 g$
Molar Mass= $100 g$
No. of moles= $\frac{100}{100} = 1$
HCl, $1 M$ Solution means $1$ mole in $1 L$ of solution
$C a C O_{3} + 2 H C l ⟶ C a {C l}_{2} + {C O}_{2} + H_{2} O$
$1$ mol $1$ mol
The ratio $\frac{m o l e}{S l i o c h r o m e t r i c C o e f f} \Rightarrow l o w e s t \to l i m i t i n g r e a g e n t C a C O_{3} ⟶ \frac{1}{1} = 1 H C l ⟶ \frac{1}{2} = 0.5 C a C O_{3} > H C l$
Thus, HCl is the limiting reagent
$2 m o l e s o f H C l ⟶ 1 m o l o f {C O}_{2} 1 m o l e o f H C l ⟶ \frac{1}{2} m o l o f {C O}_{2} ⟶ 0.5 m o l o f {C O}_{2} {C O}_{2} - M o l a r M a s s ⟶ 44 g {m o l}^{- 1} n = \frac{G i v e n M a s s}{M o l a r M a s s} 0.5 = \frac{x}{44} 0.5 \times 44 = x 22 g = x$
Weight of ${C O}_{2}$ obtained is $22 g$

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Similar questions

100 g $C a C O_{3}$ reacts with 1litre 1 N HCl. On completion of reaction how much weight of $C O_{2}$ will be obtained

100 g

C a {C O}_{3}

1

1 N

H C l

{C O}_{2}

{C O}_{2}

100

{C a C O}_{3}

H C l

1.0

C a C O_{3} (s)

0.22

C O_{2}

H C l

H C l

C a C O_{3} + 2 H C l \to C O_{2} + H_{2} O + C a C l_{2}

C a C O_{3} = 100

C O_{2} = 44

C a = 40

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