Đáp án:
\(\begin{array}{l}
a)\\
\% {m_{Ca}} = 60,3\% \\
\% {m_{CaO}} = 39,7\% \\
b)\\
{m_{F{e_3}{O_4}}} = 52,2g\\
{m_{Fe}} = 12,6g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Ca + 2{H_2}O \to Ca{(OH)_2} + {H_2}\\
CaO + {H_2}O \to Ca{(OH)_2}\\
{n_{{H_2}}} = \dfrac{{6,72}}{{22,4}} = 0,3\,mol\\
{n_{Ca}} = {n_{{H_2}}} = 0,3\,mol\\
\% {m_{Ca}} = \dfrac{{0,3 \times 40}}{{19,9}} \times 100\% = 60,3\% \\
\% {m_{CaO}} = 100 - 60,3 = 39,7\% \\
b)\\
F{e_3}{O_4} + 4{H_2} \xrightarrow{t^0} 3Fe + 4{H_2}O\\
{n_{F{e_3}{O_4}}} = \dfrac{{69,6}}{{232}} = 0,3\,mol\\
\dfrac{{{n_{F{e_3}{O_4}}}}}{1} > \dfrac{{{n_{{H_2}}}}}{4} \Rightarrow F{e_3}{O_4} \text{ dư } \\
{n_{F{e_3}{O_4}}} \text{ dư }= 0,3 - \dfrac{{0,3}}{4} = 0,225\,mol\\
{m_{F{e_3}{O_4}}} \text{ dư } = 0,225 \times 232 = 52,2g\\
{n_{Fe}} = \dfrac{{0,03}}{4} \times 3 = 0,225\,mol\\
{m_{Fe}} = 0,225 \times 56 = 12,6g
\end{array}\)