\(n_{CO_2}=1,5\left(mol\right)\\ n_{Ca\left(OH\right)_2}=1\left(mol\right)\\ T=\dfrac{2n_{Ca\left(OH\right)_2}}{n_{CO_2}}=\dfrac{2}{1,5}=1,3\) \(\rightarrow\) Tạo ra 2 muối \(CaCO_3\left(a\right),Ca\left(HCO_3\right)_2\left(b\right)\)
\(PTHH:CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\)
\(\left(mol\right)\) \(a\) \(a\) \(a\)
\(PTHH:2CO_2+Ca\left(OH\right)_2\rightarrow Ca\left(HCO_3\right)_2\)
\(\left(mol\right)\) \(2b\) \(b\) \(b\)
\(\Rightarrow\left\{{}\begin{matrix}a+2b=1,5\\a+b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,5\\b=0,5\end{matrix}\right.\)
\(a.C_{M_{Ca\left(HCO_3\right)_2}}=\dfrac{0,5}{0,5}=1\left(M\right)\)
\(b.PTHH:Ca\left(OH\right)_2+2HCl\rightarrow CaCl_2+2H_2O\)
\(\left(mol\right)\) \(1\) \(2\)
\(m_{ddHCl}=\dfrac{36,5.2.100}{25}=292\left(g\right)\)
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