\(n_{CO_2}=0,15\left(mol\right)\\ n_{Ca\left(OH\right)_2}=0,1\left(mol\right)\rightarrow n_{OH}=0,2\left(mol\right)\\ TL:\frac{n_{OH}}{n_{CO_2}}=\frac{0,2}{0,15}=1,3\)
→ Tạo ra hh 2 muối
\(n_{CaCO_3}=x;n_{Ca\left(HCO_3\right)_2}=y\)
\(PTHH:Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3+H_2O\\ PTHH:Ca\left(OH\right)_22CO_2\rightarrow Ca\left(HCO_3\right)_2\\ hpt:\left\{{}\begin{matrix}x+y=0,1\\x+2y=0,15\end{matrix}\right.\Leftrightarrow x=y=0,05\\ \rightarrow m_{kt}=0,05.100=5\left(g\right)\)
Ta có
n\(_{CO2}=\frac{3,36}{22,4}=0,15\left(mol\right)\)
n\(_{Ca\left(OH\right)_2}=0,2.0,5=0,1\left(mol\right)\)
=> n\(_{CO2}>n_{Ca\left(OH\right)2}\Rightarrow\)Tạo 2 muối
CO2+Ca(OH)2----> CaCO3 +H2O(1)
CO2+CaCO3 +H2O---->Ca(HCO3)2(2)
Theo pthh (1)n\(_{CaCO3}=n_{Ca\left(OH\right)2}=0,1mol\)
m\(_{CaCO3}=0,1.10=10\left(g\right)\)
Chúc bạn hok tốt
Nhớ tích cho mình nhé
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