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Physical Chemistry

The decomposition of a certain mass of CaCO3, gave 11.2dm3 of Co2 at STP, the mass of KOH required to completely neutralize the gas is a. 56g b. 28g c.42g d. 20g
The book shows the ans to be 56g

Profile image of Regina Phalange
7 Years agoGrade
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1 Answer

Profile image of Pooja
7 Years ago

CaCO3➡️CaO + CO2

HereCO2 produced is 11.2 ÷22.4 =0.5mole

CO2+2KOH➡️K2CO3+H2O

1 mole CO2 requires 2 mole KOH

So 0.5 mole CO2 will require 1 mole KOH

Mass of KOH required = 1× 56= 56 gram