Đáp án:
\(\begin{array}{l}
19,D\\
20,D\\
21,C\\
22,C\\
23,B\\
24,C\\
25,D\\
26,D\\
27,B\\
28,C\\
29,D\\
30,B\\
31,C\\
32,D\\
33,B\\
34,A
\end{array}\)
Giải thích các bước giải:
23,
\(\begin{array}{l}
Zn + {H_2}S{O_4} \to ZnS{O_4} + {H_2}\\
{n_{Zn}} = 0,2mol\\
\to {n_{{H_2}}} = {n_{Zn}} = 0,2mol\\
\to {V_{{H_2}}} = 4,48l
\end{array}\)
24,
\(\begin{array}{l}
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{{H_2}}} = 0,1mol\\
\to {n_{Fe}} = {n_{{H_2}}} = 0,1mol\\
\to {m_{Fe}} = 5,6g
\end{array}\)
25,
\(\begin{array}{l}
Zn + {H_2}S{O_4} \to ZnS{O_4} + {H_2}\\
{n_{Zn}} = 0,15mol\\
{n_{{H_2}S{O_4}}} = 0,1mol\\
\to {n_{Zn}} > {n_{{H_2}S{O_4}}}
\end{array}\)
Suy ra Zn dư
\(\begin{array}{l}
\to {n_{{H_2}}} = {n_{{H_2}S{O_4}}} = 0,1mol\\
\to {V_{{H_2}}} = 2,24l
\end{array}\)
26,
\(\begin{array}{l}
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{Fe}} = 1mol\\
{n_{{H_2}}} = 0,5mol\\
\to {n_{Fe}} > {n_{{H_2}}}
\end{array}\)
Suy ra Fe dư
\(\begin{array}{l}
\to {n_{HCl}} = 2{n_{{H_2}}} = 1mol\\
\to {n_{Fe}}dư= 1 - 0,5 = 0,5mol\\
\to {n_{HCl}}cần= 2{n_{Fe}}dư= 1mol
\end{array}\)
29,
\(\begin{array}{l}
2{H_2}O \to 2{H_2} + {O_2}\\
{n_{{O_2}}} = 0,25mol\\
{n_{{H_2}}} = 0,5mol\\
\to {n_{{O_2}}} = \dfrac{1}{2}{n_{{H_2}}} = 0,25mol\\
\to {n_{{H_2}O}} = {n_{{H_2}}} = 0,5mol
\end{array}\)
