Đáp án:
${V_{C{O_2}}} = 56\,\,lít$
Giải thích các bước giải:
Đun kĩ dung dịch X thu thêm $50$ gam kết tủa → Phản ứng sinh ra hai muối $CaCO_3$ và $Ca(HCO_3)_2$
$\begin{gathered} {n_{CaC{O_3}\,\,(1)}} = \frac{{150}}{{100}} = 1,5\,\,mol \hfill \\ {n_{CaC{O_3}\,\,(2)}} = \frac{{50}}{{100}} = 0,5\,\,mol \hfill \\ \end{gathered} $
Phương trình hóa học:
$C{O_2} + Ca{(OH)_2} \to CaC{O_3} + {H_2}O$
$1,5 \leftarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1,5\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,mol$
$2C{O_2} + Ca{(OH)_2} \to Ca{(HC{O_3})_2}$
$1 \leftarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,5\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,mol$
$Ca{(HC{O_3})_2}\xrightarrow{{{t^o}}}CaC{O_3} + C{O_2} + {H_2}O$
$0,5 \leftarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,5\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,mol$
$\begin{gathered} \to {n_{C{O_2}}} = 1,5 + 1 = 2,5\,\,mol \hfill \\ \to {V_{C{O_2}}} = 2,5.22,4 = 56\,\,lít \hfill \\ \end{gathered} $
