A) 56 g
B) 28 g
C) 42 g
D) 20 g
Correct Answer: A
Solution :
\[2\text{ }KOH+C{{O}_{2}}\to {{K}_{2}}C{{O}_{3}}+{{H}_{2}}O\] 1 mol 1 mol \[2\,\,\times \,\,56\text{ }gm~~\] \[22.4\,\,litre\] 11.2 litre \[C{{O}_{2}}\] will neutralize by 56 gm of KOHYou need to login to perform this action.
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