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Q.

Volume of CO₂ obtained by the complete decomposition of 9.85 g of BaCO₃ is (molecular weight of BaCO₃ = 197.3)
( ${BaCO}_{3} \to BaO + {CO}_{2}$ )

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a

0.56 L

b

1.118 L

c

2.24 L

d

0.84 L

answer is B.

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Detailed Solution

$\begin{aligned} {BaCO}_{3} \to BaO + {CO}_{2} \\ 197.3 g - \dots - 22.4 L \\ 9.85 g \dots \\ = 1.118 L \end{aligned}$

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