Giải thích các bước giải:
\(\begin{array}{l}
4.\\
2Na + {H_2}S{O_4} \to N{a_2}S{O_4} + {H_2}(1)\\
3{H_2}S{O_4} + A{l_2}{O_3} \to A{l_2}{(S{O_4})_3} + 3{H_2}O(2)\\
{n_{{H_2}S{O_4}}} = 0,2mol\\
\to {n_{{H_2}S{O_4}(2)}} = 3{n_{A{l_2}{O_3}}} = 0,15mol\\
\to {n_{{H_2}S{O_4}(1)}} = 0,2 - 0,15 = 0,05mol\\
\to {n_{Na}} = 2{n_{{H_2}S{O_4}(1)}} = 0,1mol\\
\to {m_{Na}} = 2,3g\\
7.\\
CuO + 2HCl \to CuC{l_2} + {H_2}O(1)\\
FeO + 2HCl \to FeC{l_2} + {H_2}O(2)\\
{n_{CuO}} = 0,1mol\\
{n_{FeO}} = 0,05mol\\
{n_{HCl}} = 0,24mol\\
\to {n_{HCl(1)}} = 2{n_{CuO}} = 0,2mol\\
\to {n_{HCl(2)}} = 0,04mol\\
\to {n_{CuC{l_2}}} = {n_{CuO}} = 0,1mol\\
\to {n_{FeC{l_2}}} =1/2{n_{HCl(2)}} = 0,02mol\\
\to {m_{muối}} = {m_{CuC{l_2}}} + {m_{FeC{l_2}}} = 13,5 + 2,54 = 16,04g
\end{array}\)