Ba(OH)2 + CO2 \(\rightarrow\)BaCO3 + H2O
nBa(OH)2=0,1.1=0,1(mol)
nBaCO3=\(\dfrac{19,7}{197}=0,1\left(mol\right)\)
Vì 0,1=0,1 nên tác dụng vừa đủ
Theo PTHH ta có:
nBaCO3=nCO2=0,1(mol)
VCO2=22,4.0,1=2,24(lít)
%VCO2=\(\dfrac{2,24}{4,48}.100\%=50\%\)
%VN2=100-50=50%