Đáp án:
\(\begin{array}{l}
8\,D\\
9\,C\\
10\,C
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
8)\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
{n_{{H_2}}} = \dfrac{{2,24}}{{22,4}} = 0,1\,mol\\
{n_{Fe}} = {n_{{H_2}}} = 0,1\,mol\\
m = {m_{Cu}} = 10 - 0,1 \times 56 = 4,4g\\
9)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{1}{2} = 0,5\,mol\\
{n_{HCl}} = 2{n_{{H_2}}} = 1\,mol\\
BTKL:\\
m = {m_{hh}} + {m_{HCl}} - {m_{{H_2}}} = 20 + 1 \times 36,5 - 1 = 55,5g\\
10)\\
Zn + {H_2}S{O_4} \to ZnS{O_4} + {H_2}\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
{n_{{H_2}}} = \dfrac{{8,96}}{{22,4}} = 0,4\,mol\\
{n_{{H_2}S{O_4}}} = {n_{{H_2}}} = 0,4\,mol\\
BTKL:\\
m = {m_{hh}} + {m_{{H_2}S{O_4}}} - {m_{{H_2}}} = 11,9 + 0,4 \times 98 - 0,4 \times 2 = 50,3g
\end{array}\)
