The volume of $C O_{2}$ obtained by the complete decomposition of 9.85 gm $B a C O_{3}$ at STP is :

2.24 lit.

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1.12 lit.

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0.84 lit.

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0.56 lit.

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Solution

The correct option is B 1.12 lit.
$B a C O_{3} ⟶ B a O + C O_{2}$

1 mole of $B a C O_{3}$ gives 1 mole of $C O_{2}$ .

197.34g of $B a C O_{3}$ gives 44g of $C O_{2}$ .

9.85g of $B a C O_{3}$ gives $= \frac{9.85 \times 44}{197.34} = 2.196 g o f C O_{2}$

No. of moles of $C O_{2} = \frac{2.196}{44} = 0.05 m o l e s$

$∴$ Volume of $C O_{2} = (0.05) \times 22.4 l i t r e s$

$∴$ Volume of $C O_{2} = 1.12 l i t r e s$

Hence, the correct option is $B$

Suggest Corrections

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