Đáp án:
V=11,2l
\(C{\% _{Ca{{(HC{O_3})}_2}}} = \dfrac{{0,1 \times 162}}{{140}} \times 100\% = 11,57\% \)
Giải thích các bước giải:
\(\begin{array}{l}
{n_{Ca{{(OH)}_2}}} = \dfrac{{148 \times 20}}{{100 \times 74}} = 0,4mol\\
{n_{CaC{O_3}}} = 0,3mol\\
Ca{(OH)_2} + C{O_2} \to CaC{O_3} + {H_2}O\\
{n_{Ca{{(OH)}_2}}} < {n_{CaC{O_3}}}
\end{array}\)
Suy ra có tạo ra \(Ca{(HC{O_3})_2}\)
\(\begin{array}{l}
Ca{(OH)_2} + 2C{O_2} \to Ca{(HC{O_3})_2}\\
Ca{(OH)_2} + C{O_2} \to CaC{O_3} + {H_2}O
\end{array}\)
Bảo toàn nguyên tố Ca, ta có:
\(\begin{array}{l}
{n_{Ca{{(HC{O_3})}_2}}} = {n_{Ca{{(OH)}_2}}} - {n_{CaC{O_3}}} = 0,1mol\\
\to {n_{C{O_2}}} = 2{n_{Ca{{(HC{O_3})}_2}}} + {n_{CaC{O_3}}} = 0,5mol\\
\to {V_{C{O_2}}} = 11,2l
\end{array}\)
\(\begin{array}{l}
{m_{{\rm{dd}}}} = 0,5 \times 44 + 148 - 30 = 140g\\
\to C{\% _{Ca{{(HC{O_3})}_2}}} = \dfrac{{0,1 \times 162}}{{140}} \times 100\% = 11,57\%
\end{array}\)
