Đáp án:

\(\begin{array}{l}
19,D\\
20,C\\
21,C\\
22,D\\
23,B\\
24,A\\
25,D\\
26,C
\end{array}\)

Giải thích các bước giải:

19,

\(\begin{array}{l}
Mg + 2HCl \to MgC{l_2} + {H_2}\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
Fe + 2HCl \to ZnC{l_2} + {H_2}
\end{array}\)

20, \(CuS{O_4} + Ba{(OH)_2} \to BaS{O_4} + Cu{(OH)_2}\)

21, \(Zn + MgS{O_4},Ag + HCl\)

22, \(MgS{O_4} + BaC{l_2} \to BaS{O_4} + MgC{l_2}\)

23,

\(\begin{array}{l}
Fe \to FeO \to FeC{l_2} \to FeC{l_3} \to Cl\\
2Fe + {O_2} \to 2FeO\\
FeO + 2HCl \to FeC{l_2} + {H_2}O\\
2FeC{l_2} + C{l_2} \to 2FeC{l_3}\\
2FeC{l_3} \to 2Fe + 3C{l_2}
\end{array}\)

24,

\(\begin{array}{l}
2KOH + C{O_2} \to {K_2}C{O_3} + {H_2}O\\
{n_{KOH}}du = 0,5 - 2{n_{C{O_2}}} = 0,1mol\\
{n_{{K_2}C{O_3}}} = {n_{C{O_2}}} = 0,2mol
\end{array}\)

25,

\(\begin{array}{l}
CaC{O_3} + 2HCl \to CaC{l_2} + C{O_2} + {H_2}O\\
{n_{C{O_2}}} = 0,2mol\\
 \to {n_{CaC{O_3}}} = {n_{C{O_2}}} = 0,2mol\\
 \to {m_{CaC{O_3}}} = 20g
\end{array}\)

26,

\(\begin{array}{l}
BaC{l_2} + {H_2}S{O_4} \to BaS{O_4} + 2HCl\\
{n_{BaC{l_2}}} = 0,03mol\\
{n_{{H_2}S{O_4}}} = 0,06mol\\
 \to {n_{BaC{l_2}}} < {n_{{H_2}S{O_4}}}
\end{array}\)

Suy ra \({H_2}S{O_4}\) dư 

\( \to {n_{BaS{O_4}}} = {n_{BaC{l_2}}} = 0,03mol\)