A) 1.12 L
B) 0.84 L
C) 2.24 L
D) 4.06 L
Correct Answer: A
Solution :
On decomposition\[~BaC{{O}_{3}}\] liberates \[C{{O}_{2}}\]as \[\underset{197\,g}{\mathop{BaC{{O}_{3}}}}\,\xrightarrow{{}}BaO+\underset{22.4\,\text{L}\,\text{at}\,\text{STP}}{\mathop{C{{O}_{2}}\uparrow }}\,\] \[\because \]197 g \[BaC{{O}_{3}}\] gives 22.4 L of \[\text{C}{{\text{O}}_{\text{2}}}\]at STP \[\therefore \]\[\text{9}\text{.85 g BaC}{{\text{O}}_{\text{3}}}\] will give\[\text{C}{{\text{O}}_{2}}\] \[=\frac{22.4\times 9.85}{197}=1.12\,L\]
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