A) 1.12L
B) 2.24L
C) 4.06L
D) 0.84L
Correct Answer: A
Solution :
[a] \[BaC{{O}_{3}}\xrightarrow{{}}BaO+C{{O}_{2}}\] 192 g of \[BaC{{O}_{3}}\] gives 1 mol of \[C{{O}_{2}}=22.4L\] 9.85 g of \[BaC{{O}_{3}}\] will give 0.05 mol of \[C{{O}_{2}}\] which is equal to 1.12 litre.You need to login to perform this action.
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