Em tham khảo nha :
\(\begin{array}{l}
5)\\
a)\\
F{e_2}{O_3} + 6HCl \to 2FeC{l_2} + 3{H_2}O\\
MgO + 2HCl \to MgC{l_2} + {H_2}O\\
hh:F{e_2}{O_3}(a\,mol),MgO(b\,mol)\\
\left\{ \begin{array}{l}
160a + 40b = 16\\
325a + 95b = 35,25
\end{array} \right.\\
\Rightarrow a = 0,05mol;b = 0,2mol\\
{m_{F{e_2}{O_3}}} = 0,05 \times 160 = 8g\\
\% F{e_2}{O_3} = \dfrac{8}{{16}} \times 100\% = 50\% \\
\% MgO = 100 - 50 = 50\% \\
b)\\
{n_{HCl}} = 6{n_{F{e_2}{O_3}}} + 2{n_{MgO}} = 0,7mol\\
{C_{{M_{HCl}}}} = \dfrac{{0,7}}{{0,3}} = \frac{7}{3}M\\
6)\\
CaC{O_3} \to CaO + C{O_2}\\
{m_{CaC{O_3}}} = \dfrac{{3000 \times 85}}{{100}} = 2550kg = 2550000g\\
{n_{CaC{O_3}}} = \dfrac{{2550000}}{{100}} = 25500mol\\
{n_{CaO}} = {n_{CaC{O_3}}} = 25500mol\\
{m_{CaO}} = 25500 \times 56 = 1428000g = 1428kg\\
{m_{Ca{O_{td}}}} = \dfrac{{1428 \times 90}}{{100}} = 1285,2kg
\end{array}\)
