Giải thích các bước giải:
7.
\(\begin{array}{l}
Na + {H_2}O \to NaOH + \dfrac{1}{2}{H_2}(1)\\
2NaOH + A{l_2}{O_3} \to 2NaAl{O_2} + {H_2}O(2)\\
Fe + 2HCl \to FeC{l_2} + {H_2}(3)
\end{array}\)
\(\begin{array}{l}
a)\\
{n_{{H_2}(1)}} = {n_{{H_2}(3)}} = 0,3mol\\
\to {n_{Na}} = {n_{NaOH}} = 2{n_{{H_2}(1)}} = 0,6mol\\
\to {m_{Na}} = 13,8g\\
{m_{Fe}} = 16,8g \to {n_{Fe}} = 0,3mol\\
\to {m_{A{l_2}{O_3}}} = 51 - (13,8 + 16,8) = 20,4g\\
\to {n_{A{l_2}{O_3}}} = 0,2mol\\
\to {n_{NaOH(2)}}dư\\
b)\\
{m_{NaOH}} = 24g\\
{n_{NaAl{O_2}}} = 2{n_{A{l_2}{O_3}}} = 0,4mol\\
\to {m_{NaAl{O_2}}} = 32,8g\\
{n_{FeC{l_2}}} = {n_{{H_2}(3)}} = 0,3mol\\
\to {m_{FeC{l_2}}} = 38,1g\\
\to {m_{{\rm{dd}}A}} = {m_{Na}} + {m_{A{l_2}{O_3}}} + {m_{{H_2}O}} - {m_{{H_2}}} = 433,6g\\
{n_{HCl}} = 2{n_{{H_2}(3)}} = 0,6mol \to {m_{HCl}} = 21,9g\\
\to {m_{{\rm{dd}}HCl}} = \dfrac{{21,9 \times 100}}{{25}} = 87,6g\\
\to {m_{{\rm{dd}}B}} = {m_{Fe}} + {m_{{\rm{dd}}HCl}} - {m_{{H_2}}} = 103,8g\\
\to C{\% _{NaOH}} = \dfrac{{24}}{{433,6}} \times 100\% = 5,54\% \\
\to C{\% _{NaAl{O_2}}} = \dfrac{{32,8}}{{433,6}} \times 100\% = 7,56\% \\
\to C{\% _{FeC{l_2}}} = \dfrac{{38,1}}{{103,8}} \times 100\% = 36,71\%
\end{array}\)
