Question

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​

Answer 1

Answer:

Given :-

➡️ Volume of CO2 = 11.2 dm3

➡ 1 mole of CO2 = 22.4 dm3

➡ 44 gm of of CO2 = 22.4 dm3

➡ x gm of CO2 = 11.2 dm3 .

➡ x = 11.4*44/22.4 =22gm

neutralisation rxn is :

koh + co2 ➡️ KHCO3

KOH required for neutralisation of 22 gm of of CO2:= 56*22/44 =28 gm

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