Đáp án:
\(\begin{array}{l}
a)\\
\% {n_{Fe}} = 44\% \\
{V_{{\rm{dd}}HCl}} = 500\,ml\\
b)\\
{m_{F{e_2}{O_3}}} = 13,33g\\
{m_{Fe}} = 9,33g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
hh:Fe(a\,mol),Mg(b\,mol)\\
\left\{ \begin{array}{l}
56a + 24b = 9,52\\
127a + 95b = 27,27
\end{array} \right.\\
\Rightarrow a = 0,11;b = 0,14\\
\% {n_{Fe}} = \dfrac{{0,11}}{{0,11 + 0,14}} \times 100\% = 44\% \\
{n_{HCl}} = 2{n_{Fe}} + 2{n_{Mg}} = 0,11 \times 2 + 0,14 \times 2 = 0,5\,mol\\
{V_{{\rm{dd}}HCl}} = \dfrac{{0,5}}{1} = 0,5l = 500\,ml\\
b)\\
F{e_2}{O_3} + 3{H_2} \to 2Fe + 3{H_2}O\\
{n_{{H_2}}} = {n_{ Fe}} + {n_{Mg}} = 0,11 + 0,14 = 0,25\,mol\\
{n_{F{e_2}{O_3}}} = 0,25 \times \dfrac{1}{3} = \dfrac{1}{{12}}\,mol\\
{m_{F{e_2}{O_3}}} = \dfrac{1}{{12}} \times 160 = 13,33g\\
{n_{Fe}} = 2{n_{F{e_2}{O_3}}} = \dfrac{1}{6}\,mol\\
{m_{Fe}} = \dfrac{1}{6} \times 56 = 9,33g
\end{array}\)
