Đáp án:

Bạn tham khảo lời giải ở dưới nhé!!

Giải thích các bước giải:

1,

\(\begin{array}{l}
F{e_2}{O_3} + 6HCl \to 2FeC{l_3} + 3{H_2}O\\
{n_{F{e_2}{O_3}}} = 0,06mol\\
 \to {n_{HCl}} = 6{n_{F{e_2}{O_3}}} = 0,36mol\\
 \to {m_{HCl}} = 13,14g\\
 \to x = C{\% _{HCl}} = \dfrac{{13,14}}{{150}} \times 100\%  = 8,76\% \\
 \to {n_{FeC{l_3}}} = 2{n_{F{e_2}{O_3}}} = 0,12mol\\
 \to {m_{FeC{l_3}}} = 19,5g
\end{array}\)

2,

\(\begin{array}{l}
Fe + 2HCl \to FeC{l_2} + {H_2}\\
MgO + 2HCl \to MgC{l_2} + {H_2}\\
{n_{{H_2}}} = 0,3mol\\
 \to {n_{Fe}} = {n_{{H_2}}} = 0,3mol\\
 \to {m_{Fe}} = 16,8g\\
 \to {m_{MgO}} = 5,9g\\
 \to \% {m_{Fe}} = \dfrac{{16,8}}{{22,7}} \times 100\%  = 74\% \\
 \to \% {m_{MgO}} = 36\% \\
 \to {n_{MgO}} = 0,1475mol\\
 \to {n_{HCl}} = 2{n_{Fe}} + 2{n_{MgO}} = 0,895mol\\
 \to C{M_{HCl}} = \dfrac{{0,895}}{{0,25}} = 3,58M
\end{array}\)

3,

\(\begin{array}{l}
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
CuO + 2HCl \to CuC{l_2} + {H_2}O\\
{n_{{H_2}}} = 0,15mol\\
 \to {n_{Al}} = \dfrac{2}{3}{n_{{H_2}}} = 0,1mol\\
 \to {m_{Al}} = 2,7g\\
 \to {m_{CuO}} = 22,7 - 2,7 = 20g\\
 \to {n_{CuO}} = 0,25mol\\
 \to \% {m_{Al}} = \dfrac{{2,7}}{{22,7}} \times 100\%  = 11,89\% \\
 \to \% {m_{CuO}} = 88,11\% \\
 \to {n_{HCl}} = 3{n_{Al}} + 2{n_{CuO}} = 0,8mol\\
 \to {V_{HCl}} = \dfrac{{0,8}}{2} = 0,4l
\end{array}\)

4,

\(\begin{array}{l}
N{a_2}O + {H_2}O \to 2NaOH\\
{n_{N{a_2}O}} = 0,2mol\\
 \to {n_{NaOH}} = 2{n_{N{a_2}O}} = 0,4mol\\
 \to C{M_{NaOH}} = \dfrac{{0,4}}{{0,8}} = 0,5M\\
2NaOH + {H_2}S{O_4} \to N{a_2}S{O_4} + 2{H_2}O\\
{n_{{H_2}S{O_4}}} = \dfrac{1}{2}{n_{NaOH}} = 0,2mol\\
 \to {V_{{H_2}S{O_4}}} = \dfrac{n}{{CM}} = \dfrac{{0,2}}{{0,5}} = 0,4l
\end{array}\)