Đáp án:
\(\begin{array}{l}
27,C\\
29,A\\
30,A\\
31,A\\
33,A\\
35,A\\
36,A\\
37,A\\
38,A\\
39,A\\
40,D\\
41,D\\
42,D\\
43,B\\
44,C\\
45,D
\end{array}\)
Giải thích các bước giải:
27,
\(\begin{array}{l}
MgO + {H_2}S{O_4} \to MgS{O_4} + {H_2}O\\
F{e_2}{O_3} + 3{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3{H_2}O\\
CuO + {H_2}S{O_4} \to CuS{O_4} + {H_2}O\\
{K_2}O + {H_2}S{O_4} \to {K_2}S{O_4} + {H_2}O
\end{array}\)
29,
\(\begin{array}{l}
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
{n_{Fe}} = 0,15mol\\
\to {n_{{H_2}}} = {n_{Fe}} = 0,15mol\\
\to {V_{{H_2}}} = 3,36l
\end{array}\)
31,
\(\begin{array}{l}
Fe{S_2} \to 2{H_2}S{O_4}\\
{n_{Fe{S_2}}} = 0,5mol\\
\to {n_{{H_2}S{O_4}}} = 2{n_{Fe{S_2}}} = 1mol\\
\to {m_{{H_2}S{O_4}}} = 98kg
\end{array}\)
33,
\(\begin{array}{l}
Zn + {H_2}S{O_4} \to ZnS{O_4} + {H_2}\\
{n_{{H_2}}} = 0,1mol\\
\to {n_{Zn}} = {n_{{H_2}}} = 0,1mol\\
\to \% {m_{Zn}} = \dfrac{{0,1 \times 65}}{{10,5}} \times 100\% = 61,9\% \\
\to \% {m_{Cu}} = 38,1\%
\end{array}\)
36,\(S{O_2} + {H_2}O \to {H_2}S{O_3}\)
37, \(Fe + 2HCl \to FeC{l_2} + {H_2}\)
