Đáp án:

 \(\begin{array}{l}
1.\\
C{\% _{ZnS{O_4}}} = 32,2\% 
\end{array}\)

\(\begin{array}{l}
6.\\
a.\\
\% {m_{Al}} = 49,09\% \\
\% {m_{Fe}} = 50,91\% \\
b.\\
{V_{{\rm{dd}}HCl}} = 92,59ml = 0,09l\\
c.\\
C{\% _{AlC{l_3}}} = 12,7\% \\
C{\% _{FeC{l_2}}} = 6,04\% 
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
1.\\
Zn + {H_2}S{O_4} \to ZnS{O_4} + {H_2}\\
{n_{Zn}} = 0,4mol\\
 \to {n_{ZnS{O_4}}} = {n_{Zn}} = 0,4mol\\
 \to {m_{ZnS{O_4}}} = 64,4g\\
 \to C{\% _{ZnS{O_4}}} = \dfrac{{64,4}}{{200}} \times 100\%  = 32,2\% 
\end{array}\)

 \(\begin{array}{l}
6.\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{{H_2}}} = 0,2mol
\end{array}\)

Gọi a và b lần lượt là số mol của Al và Fe

\(\begin{array}{l}
\left\{ \begin{array}{l}
27a + 56b = 5,5\\
\dfrac{3}{2}a + b = 0,2
\end{array} \right. \to \left\{ \begin{array}{l}
a = 0,1\\
b = 0,05
\end{array} \right.\\
 \to {n_{Al}} = 0,1mol\\
 \to {n_{Fe}} = 0,05mol\\
a.\\
\% {m_{Al}} = \dfrac{{0,1 \times 27}}{{5,5}} \times 100\%  = 49,09\% \\
\% {m_{Fe}} = 100\%  - 49,09\%  = 50,91\% \\
b.\\
{n_{HCl}} = 3{n_{Al}} + 2{n_{Fe}} = 0,4mol\\
 \to {m_{HCl}} = 14,6g\\
 \to {m_{{\rm{dd}}HCl}} = \dfrac{{14,6}}{{14,6\% }} \times 100\%  = 100g\\
 \to {V_{{\rm{dd}}HCl}} = \dfrac{{100}}{{1,08}} = 92,59ml = 0,09l\\
c.\\
{n_{AlC{l_3}}} = {n_{Al}} = 0,1mol\\
 \to {m_{AlC{l_3}}} = 13,35g\\
{n_{FeC{l_2}}} = {n_{Fe}} = 0,05mol\\
 \to {m_{FeC{l_2}}} = 6,35g\\
{m_{{\rm{dd}}}} = {m_{hh}} + {m_{{\rm{dd}}HCl}} - {m_{{H_2}}} = 105,1g\\
 \to C{\% _{AlC{l_3}}} = \dfrac{{13,35}}{{105,1}} \times 100\%  = 12,7\% \\
 \to C{\% _{FeC{l_2}}} = \dfrac{{6,35}}{{105,1}} \times 100\%  = 6,04\% 
\end{array}\)