Đáp án:
\(\begin{array}{l}
b)\\
\% {m_{F{e_2}{O_3}}} = 44,44\% \\
\% {m_{MgO}} = 55,56\% \\
c)\\
{V_{{\rm{dd}}HCl}} = 1,92l
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
F{e_2}{O_3} + 6HCl \to 2FeC{l_3} + 3{H_2}O\\
MgO + 2HCl \to MgC{l_2} + {H_2}O\\
b)\\
hh:F{e_2}{O_3}(a\,mol),MgO(b\,mol)\\
\left\{ \begin{array}{l}
160a + 40b = 21,6\\
162,5 \times 2a + 95b = 48
\end{array} \right.\\
\Rightarrow a = 0,06;b = 0,3\\
\% {m_{F{e_2}{O_3}}} = \dfrac{{0,06 \times 160}}{{21,6}} \times 100\% = 44,44\% \\
\% {m_{MgO}} = 100 - 44,44 = 55,56\% \\
c)\\
{n_{HCl}} = 0,06 \times 6 + 0,3 \times 2 = 0,96\,mol\\
{V_{{\rm{dd}}HCl}} = \dfrac{{0,96}}{{0,5}} = 1,92l
\end{array}\)
