text: # The decomposition of a certain mass of CaCO3 gave 11.2 dm3 of CO2 gas at STP. The mass of KOH required to completely neutralise the gas is (A) 56 g (B) 28 g (C) 42 g (D) 20 g
  
  R Shyam Sundar
  15 Points
  4 years ago
  For complete neutralisation,
  2KOH+CO2 gives K2CO3 + H2O
  For 11.2dm^3 i.e.,11.2L(half mole) we need 1mole KOH.
  So weight of KOH is 56g
  
  Vikas TU
  14149 Points
  4 years ago
  Dear student
  Weight of 11.2 dm^3 of CO2 gas at STP = 44/2 = 22 gram
  KOH + CO2 -> KHCO3
  KOH required = 56/44  * 22 = 28 gram
  B is correct ans.
  Hope this helps
  Good Luck