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