A) 56 g
B) 28 g
C) 42 g
D) 20 g
Correct Answer: B
Solution :
Weight of \[11.2\text{ }d{{m}^{3}}\] of \[C{{O}_{2}}\] gas at STP = 44/2 = 22g. \[\underset{56\,g}{\mathop{KOH}}\,+\underset{44\,g}{\mathop{C{{O}_{2}}}}\,\xrightarrow{{}}KHC{{O}_{3}}\] KOH required for complete neutralization requires of 22 g of \[C{{O}_{2}}\] is \[=\frac{56}{44}\times 22=28\,g\]You need to login to perform this action.
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