Question

The decomposition of a certain mass of CaCO 3 ​ gave 11.2 dm 3 of CO 2 ​ gas at S.T.P. The mass of KOH required to completely neutralise the gas is :​

Answer 1

$\huge\mathcal\colorbox{lavender}{{\color{b}{✿Yøur-Añswer♡}}}$

$\large\bf{\underline{\red{illusion}}}$

➡️ 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..