A) \[C{{H}_{4}}\]
B) \[{{C}_{2}}{{H}_{6}}\]
C) \[C{{O}_{2}}\]
D) \[Xe\]
Correct Answer: C
Solution :
requires of \[22g\]of \[C{{O}_{2}}\] is \[=\frac{56}{44}\times 22=28g\]
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