Đáp án:
Bạn tham khảo lời giải ở dưới nhé !
Giải thích các bước giải:
Bài 1:
\(\begin{array}{l}
Na + \dfrac{1}{2}C{l_2} \to NaCl\\
2NaCl + 2{H_2}O \to 2NaOH + {H_2} + C{l_2}\\
2NaOH + A{l_2}{O_3} \to 2NaAl{O_2} + {H_2}O\\
NaAl{O_2} + 2{H_2}O + C{O_2} \to Al{(OH)_3} + NaHC{O_3}
\end{array}\)
\(\begin{array}{l}
2Al{(OH)_3} \to A{l_2}{O_3} + 3{H_2}O\\
2A{l_2}{O_3} \to 4Al + 3{O_2}\\
2Al + 3Fe{(N{O_3})_2} \to 3Fe + 2Al{(N{O_3})_3}\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
FeC{l_2} + Mg \to MgC{l_2} + Fe\\
MgC{l_2} \to Mg + C{l_2}
\end{array}\)
Bài 2:
\(\begin{array}{l}
Na + {H_2}O \to NaOH + \dfrac{1}{2}{H_2}\\
{n_{Na}} = 1mol\\
\to {n_{NaOH}} = {n_{Na}} = 1mol\\
\to C{M_{NaOH}} = \dfrac{1}{1} = 1M\\
2Al + 2NaOH + 2{H_2}O \to 2NaAl{O_2} + 3{H_2}\\
{n_{NaOH}} = 0,2mol\\
\to {n_{Al}} = {n_{NaOH}} = 0,2mol\\
\to {m_{Al}} = 5,4g\\
\to {m_{Mg}} = 10,2 - 5,4 = 4,8g\\
\to \% {m_{Al}} = \dfrac{{5,4}}{{10,2}} \times 100\% = 52,94\% \\
\to \% {m_{Mg}} = 100\% - 52,94\% = 47,06\% \\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
{n_{Al}} = 0,2mol\\
{n_{Mg}} = 0,2mol\\
\to {n_{HCl}} = 3{n_{Al}} + 2{n_{Mg}} = 1mol\\
\to {V_{HCl}} = \dfrac{1}{4} = 0,25l
\end{array}\)
