Đáp án:
$C_{M\ dd\ Ca(HCO_3)_2}=\dfrac{1}{30}(M)$
Giải thích các bước giải:
$n_{CO_2}=\dfrac{16,8}{22,4}=0,75(mol)\\ n_{Ca(OH)_2}=9.0,05=0,45(mol)\\ T=\dfrac{n_{CO_2}}{n_{Ca(OH)_2}}=\dfrac{0,75}{0,45}=1,67\\ 1<T<2\to \text{Tạo 2 muối}\\ Đặt\ \begin{cases}n_{CaCO_3}=x(mol)\\n_{Ca(HCO_3)_2}=y(mol)\end{cases}\\ \xrightarrow{\text{BTNT C}}\ n_{CaCO_3}+2.n_{Ca(HCO_3)_2}=n_{CO_2}\\ \to x+2y=0,75\ (1)\\ \xrightarrow{\text{BTNT Ca}}\ n_{CaCO_3}+n_{Ca(HCO_3)_2}=n_{Ca(OH)_2}\\ \to x+y=0,45\ (2)\\ \xrightarrow{\text{Từ (1),(2)}}\begin{cases}x=0,15\\y=0,3\end{cases}\\ \to C_{M\ dd\ Ca(HCO_3)_2}=\dfrac{0,3}{9}=\dfrac{1}{30}(M)$
