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 \[\underset{197}{\mathop{BaC{{O}_{3}}}}\,\xrightarrow{{}}BaO+\underset{22.4\,L\,at\,STP}{\mathop{C{{O}_{2}}\uparrow }}\,\] \[\because \] \[197\,g\,BaC{{O}_{3}}\] gives \[=22.4\,L\] of \[C{{O}_{2}}\] at STP \[\because \] \[9.85\,g\,BaC{{O}_{3}}\] will give \[=\frac{22.4\times 9.85}{197}=1.12\,L\]You need to login to perform this action.
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