A) 1.12L
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 \[\underset{197\,\,g}{\mathop{BaC{{O}_{3}}}}\,\xrightarrow{{}}BaO+\underset{22.4\,\,\,L\,\,\,at\,\,\,STP}{\mathop{C{{O}_{2}}\uparrow }}\,\] \[197\,g\,\,BaC{{O}_{3}}\] gives 22.4 L of \[C{{O}_{2}}\] at STP \[\therefore \,\,9.85\,\,g\,BaC{{O}_{3}}\] will give \[C{{O}_{2}}\] \[=\frac{22.4\times 9.85}{197}\] = 1.12 L
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