Đáp án:

\(\begin{array}{l}
a)\\
\% {m_{Ca}} = 55,56\% \\
\% {m_{Ca{C_2}}} = 44,44\% \\
b)\\
m = 14,4g
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
Ca + 2{H_2}O \to Ca{(OH)_2} + {H_2}\\
Ca{C_2} + 2{H_2}O \to Ca{(OH)_2} + {C_2}{H_2}\\
hh\,C:{H_2}(a\,mol),{C_2}{H_2}(b\,mol)\\
{M_C} = 5 \times 2 = 10g/mol\\
\text{ Giả sử số mol hỗn hợp là 1 mol }\\
{m_C} = 10 \times 1 = 10g\\
\left\{ \begin{array}{l}
a + b = 1\\
2a + 26b = 10
\end{array} \right.\\
 \Rightarrow a = \dfrac{2}{3};b = \dfrac{1}{3}\\
{n_{Ca}} = {n_{{H_2}}} = \dfrac{2}{3}\,mol\\
{n_{Ca{C_2}}} = {n_{{C_2}{H_2}}} = \dfrac{1}{3}\,mol\\
\% {m_{Ca}} = \dfrac{{\frac{2}{3} \times 40}}{{\frac{2}{3} \times 40 + \dfrac{1}{3} \times 64}} \times 100\%  = 55,56\% \\
\% {m_{Ca{C_2}}} = 100 - 55,56 = 44,44\% \\
b)\\
{n_{hh}} = \dfrac{{6,72}}{{22,4}} = 0,3\,mol\\
{n_{Ca{C_2}}} = {n_{{C_2}{H_2}}} = 0,3 \times \dfrac{1}{3} = 0,1\,mol\\
{n_{Ca}} = {n_{{H_2}}} = 0,3 - 0,1 = 0,2\,mol\\
m = 0,1 \times 64 + 0,2 \times 40 = 14,4g
\end{array}\)