Đáp án:

\(\begin{array}{l}
\% {m_C} = 88,89\% \\
\% {m_H} = 11,11\% 
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
Ca{(OH)_2} + C{O_2} \to CaC{O_3} + {H_2}O(1)\\
Ca{(OH)_2} + 2C{O_2} \to Ca{(HC{O_3})_2}(2)\\
Ca{(HC{O_3})_2} \xrightarrow{t^0} CaC{O_3} + C{O_2} + {H_2}O(3)\\
{n_{CaC{O_3}(3)}} = \dfrac{2}{{100}} = 0,02\,mol\\
{n_{Ca{{(HC{O_3})}_2}}} = {n_{CaC{O_3}(3)}} = 0,02\,mol\\
{n_{C{O_2}(2)}} = 2{n_{Ca{{(HC{O_3})}_2}}} = 0,04\,mol\\
{n_{CaC{O_3}(1)}} = \dfrac{4}{{100}} = 0,04\,mol\\
{n_{C{O_2}(1)}} = {n_{CaC{O_3}(1)}} = 0,04\,mol\\
{n_{C{O_2}}} = 0,04 + 0,04 = 0,08\,mol\\
{n_C} = {n_{C{O_2}}} = 0,08\,mol\\
{m_{C{O_2}}} + {m_{{H_2}O}} = 4,6g \Rightarrow {m_{{H_2}O}} = 4,6 - 0,08 \times 44 = 1,08g\\
{n_{{H_2}O}} = \dfrac{{1,08}}{{18}} = 0,06\,mol\\
{n_H} = 2{n_{{H_2}O}} = 0,12\,mol\\
{m_C} + {m_H} = 0,08 \times 12 + 0,12 = 1,08g = {m_A}\\
 \Rightarrow \\
\% {m_C} = \dfrac{{0,08 \times 12}}{{1,08}} \times 100\%  = 88,89\% \\
\% {m_H} = 100 - 88,89 = 11,11\% 
\end{array}\)