Xét bình 1:
${m_{{H_2}O}} = {m_{Binh1\,\,\operatorname{tang} }} = 0.45g\, \to {n_{{H_2}O}} = 0.025\,mol \to {n_H} = 0.05\,mol$
Xét bình 2: Dẫn CO2 qua 0.02 mol Ca(OH)2 thu được 0.01 mol kết tủa, có 2 trường hợp xảy ra
TH1: $Ca{(OH)_{2\,}}du$
$\begin{gathered} C{O_2} + Ca{(OH)_2} \to CaC{O_3} + {H_2}O \hfill \\ 0.01\,\,\,\, \leftarrow \,\,\,\,\,0.01\,\,\,\,\,\,\,\, \leftarrow \,0.01\,\,\,\,\,(mol) \hfill \\ \end{gathered} $
Bt C: $ \to {n_{C\,trongA}} = {n_{C{O_2}}} = 0.01\,mol$
$\to A\,chua\left\{ \begin{gathered} 0.05\,mol\,H \hfill \\ 0.01\,mol\,C \hfill \\ \frac{{0.73 - 0.05 \times 1 - 0.01 \times 12}}{{16}} = 0.035\,mol\,O \hfill \\ \end{gathered} \right. \to \left\{ \begin{gathered} \% H = 6.85\% \hfill \\ \% C = 16.44\% \hfill \\ \% O = 76.71\% \hfill \\ \end{gathered} \right.$
TH2:$C{O_2}\,du$
$\begin{gathered} C{O_2} + Ca{(OH)_2} \to CaC{O_3} + {H_2}O \hfill \\ 0.02\,\,\,\, \leftarrow \,\,\,\,\,0.02\,\,\,\,\,\,\,\,\, \to \,0.02\,\,\,\,\,(mol) \hfill \\ C{O_{{2_{du}}}} + CaC{O_3} + {H_2}O \to Ca{(HC{O_3})_2}_{} \hfill \\ (x - 0.02) \to (x - 0.02)\,\,\,\,(mol) \hfill \\ {n_{CaC{O_3}\,con}} = 0.02 - (x - 0.02) = 0.01 \to x = 0.03 \hfill \\ \end{gathered} $
Bt C: $ \to {n_{C\,trongA}} = {n_{C{O_2}}} = 0.03\,mol$
$\to A\,chua\left\{ \begin{gathered} 0.05\,mol\,H \hfill \\ 0.03\,mol\,C \hfill \\ \frac{{0.73 - 0.05 \times 1 - 0.03 \times 12}}{{16}} = 0.02\,mol\,O \hfill \\ \end{gathered} \right. \to \left\{ \begin{gathered} \% H = 6.85\% \hfill \\ \% C = 49.32\% \hfill \\ \% O = 43.84\% \hfill \\ \end{gathered} \right.$
