Volume of CO₂ obtained by the complete decomposition of 9.85 g BaCO₃ is :

6 2000 AIPMT Some Basic Concepts of ChemistryStoichiometry Easy
Volume of $CO_2$ obtained by the complete decomposition of 9.85 g $BaCO_3$ is :
A 2.24 lit.
B 1.12 lit.
C 0.84 lit.
D 0.56 lit.

Detailed Solution

$BaCO_3 \xrightarrow{\Delta} BaO + CO_2$
Molar mass of $BaCO_3 = 137 + 12 + 48 = 197$ g/mol
Moles of $BaCO_3 = \dfrac{9.85}{197} = 0.05$ mol, so 0.05 mol of $CO_2$ is formed.
Volume at STP $= 0.05 \times 22.4 = 1.12$ L

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