\(n_{CO2}=\frac{6.72}{22.4}=0.3mol\)
PTHH:
\(CO_2+Ba\left(OH\right)_2\rightarrow BaCO_3+H_2O\)
0.3 mol \(\rightarrow\) 0.3 mol \(\rightarrow\) 0.3 mol
\(m_{BaCO3}=0.3\cdot197=59.1\left(g\right)\)
\(C_M_{Ba\left(OH\right)2}=\frac{0.3}{0.6}=0.5\left(M\right)\)
nCO2 = 0.3 mol
Ba(OH)2 + CO2 => BaCO3 + H2O
mBaCO3 = 0.3*197 = 59.1 g
CM Ba(OH)2 = 0.3/0.6 = 0.5M