PTHH: \(C+O_2\xrightarrow[]{t^o}CO_2\)
Ta có: \(\left\{{}\begin{matrix}n_{CO_2}=n_C=\dfrac{2,4}{12}=0,2\left(mol\right)\\n_{Ca\left(OH\right)_2}=\dfrac{150\cdot7,4\%}{74}=0,15\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) Tạo 2 muối
PTHH: \(Ca\left(OH\right)_2+2CO_2\rightarrow Ca\left(HCO_3\right)_2\)
a_____2a_____________a (mol)
\(Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3+H_2O\)
b_____b_________b (mol)
Ta lập HPT: \(\left\{{}\begin{matrix}2a+b=0,2\\a+b=0,15\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,05\\b=0,1\end{matrix}\right.\)
\(\Rightarrow C\%_{Ca\left(HCO_3\right)_2}=\dfrac{0,05\cdot162}{0,2\cdot44+150-0,1\cdot100}\cdot100\%\approx5,44\%\)
