Volume of CO2 obtained by the complete decomposition of 9.85 g of BaCO3 is (molecular weight of BaCO3 = 197.3)(BaCO3→BaO+CO2)

BaCO3→BaO+CO2 197.3g−…−22.4L 9.85g… =1.118L

BaCO3→BaO+CO2 197.3g−…−22.4L 9.85g… =1.118L

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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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2.24 L

1.118 L

0.84 L

0.56 L

answer is B.

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

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