CaCO3 ----to--> CaO+ CO2
nCaCO3=0.75 mol
Theo Pt nCO2=nCaCO3=0.75 mol
=> VCO2=0.75*22.4*80%=13.44 lít =0.6 mol
b) Ba(OH)2+ CO2 ------> BaCO3+ H2O
nCO2=0.6 mol
Theo Pt=> nBa(OH)2=nBaCO3=nCO2=0.6 mol
=>mdd Ba(OH)2=(0.6*171)/17.1%=600 g
mdd=600+0.6*44=626.4 g
=>C%BaCO3=(0.6*197*100)/624.4=18.93%