Đáp án:
3,75g và 1,0125g
Giải thích các bước giải:
\(\begin{array}{l}
CaO + {H_2}O \to Ca{(OH)_2}\\
nCaO = \dfrac{{3,92}}{{56}} = 0,07\,mol\\
nCa{(OH)_2} = nCaO = 0,07\,mol\\
CMCa{(OH)_2} = \dfrac{{0,07}}{{0,8}} = 0,0875M\\
nCa{(OH)_2} = 0,0875 \times 0,5 = 0,04375\,mol\\
nC{O_2} = \dfrac{{1,12}}{{22,4}} = 0,05\,mol\\
T = \dfrac{{nCa{{(OH)}_2}}}{{nC{O_2}}} = 0,875
\end{array}\)
0,5<T<0,75=> tạo 2 muối
\(\begin{array}{l}
Ca{(OH)_2} + C{O_2} \to CaC{O_3} + {H_2}O\\
0,04375\,\,\,\,\,\,\,\,\,\,0,05\,\,\,\,\,\,\,0,04375\\
CaC{O_3} + C{O_2} + {H_2}O \to Ca{(HC{O_3})_2}\\
0,00625\,\,\,\,\,0,00625\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,00625\\
mCaC{O_3} = (0,04375 - 0,00625) \times 100 = 3,75g\\
mCa{(HC{O_3})_2} = 0,00625 \times 162 = 1,0125g
\end{array}\)
