n\(_{CO}\)=6,72/22,4=0,3mol
PTPU
\(Fe_2O_3+3CO->2Fe+3CO_2\)
0,1.............0,3...........0,2............0,3(mol)
\(m_{Ca\left(OH\right)_2}=\dfrac{7,4.200}{100}=14,8g\)
\(n_{Ca\left(OH\right)_2}\)=14,8/74=0,2mol
n\(_{CO_2}\)=0,3mol
PTPU
\(Ca\left(OH\right)_2+CO_2->CaCO_3+H_2O\)
x.....................x....................x...............x(mol)
\(Ca\left(OH\right)_2+2CO_2->Ca\left(HCO_3\right)_2\)
y........................2y....................y(mol)
n\(_{Ca\left(OH\right)_2}\)=x+y=0,2mol
n\(_{CO_2}\)=x+2y=0,3mol
=>x=0,1mol;y=0,1mol
n\(_{CaCO_3}\)=0,1mol
m\(_{CaCO_3}\)=0,1.100=10g
n\(_{Ca\left(HCO_3\right)_2}\)=0,1mol
m\(_{Ca\left(HCO_3\right)_2}\)=0,1.162=16,2g
