\(n_{CO_2}=\frac{6,72}{22,4}=0,3\left(mol\right)\\ n_{Ca\left(OH\right)_2}=1.0,25=0,25\left(mol\right)\\ \rightarrow n_{OH}=0,5\left(mol\right)\\ TL:\frac{n_{OH}}{n_{CO_2}}=\frac{0,5}{0,3}=1,6\)
→ Tạo ra hh 2 muối
\(n_{CaCO_3}=x;n_{Ca\left(HCO_3\right)_2}=y\)
\(PTHH:Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3+H_2O\)
\(PTHH:Ca\left(OH\right)_2+2CO_2\rightarrow Ca\left(HCO_3\right)_2\)
\(\Rightarrow hpt:\left\{{}\begin{matrix}x+y=0,25\\x+2y=0,3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,05\end{matrix}\right.\)
\(m_{spu}=100.0,2+0,05.162=28,1\left(g\right)\)
PTHH: Ca(OH)2 + CO2 -> CaCO3 + H2
0.25 0.3 0.25
PTHH : CaCO3 + CO2 + H2O -> Ca(HCO3)2
0.05 0.05
-> mCaCO3 SPU = 20g
=> mdd giảm = mCaCO3 - mCO2 = 6.8 g ( thêm vào 13.2 g CO2 nhưng bị kết tủa ra hết 20 => dd giảm)
