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) 56g

b) 28g

c) 42g

d)20g

Asked by Balbir | 21st Jun, 2019, 09:57: PM

Expert Answer:

Given:
 
 
Volume of CO2 = 11.2 dm3 
 
We have,
 
1 mole of CO2 = 22.4 dm3 
 
44 gm of of CO2 = 22.4 dm3 
 
x gm of CO2 = 11.2 dm3 
 
x = 
  equals fraction numerator 11.2 cross times 44 over denominator 22.4 end fraction

equals 22 space gm
 
The neutralisation reaction is as,
 
KOH space space plus space space space CO subscript 2 space space space rightwards arrow space space KHCO subscript 3
56 space gm space space space space space 44 space gm space space space space space
56 gm KOH required for neutralisation of 44 gm of of CO2 
 
KOH required for neutralisation of 22 gm of of CO2 
equals fraction numerator 56 cross times 22 over denominator 44 end fraction

equals space 28 space gm
 
28 gm of KOH is required for complete neutralisation of 22 gm of of CO2.

Answered by Varsha | 23rd Jun, 2019, 10:32: PM