Em tham khảo nha:
1)
\(\begin{array}{l}
a)\\
Zn + {H_2}S{O_4} \to ZnS{O_4} + {H_2}\\
ZnS{O_4} + BaC{l_2} \to ZnC{l_2} + BaS{O_4}\\
ZnC{l_2} + 2AgN{O_3} \to Zn{(N{O_3})_2} + 2AgCl\\
Zn{(N{O_3})_2} + 2NaOH \to Zn{(OH)_2} + 2NaN{O_3}\\
Zn{(OH)_2} \to ZnO + {H_2}O\\
b)\\
Cu + 2{H_2}S{O_4} \to CuS{O_4} + S{O_2} + 2{H_2}O\\
CuS{O_4} + BaC{l_2} \to CuC{l_2} + BaS{O_4}\\
CuC{l_2} + 2NaOH \to Cu{(OH)_2} + 2NaCl\\
Cu{(OH)_2} + 2HN{O_3} \to Cu{(N{O_3})_2} + 2{H_2}O\\
Cu{(OH)_2} \to CuO + {H_2}O\\
CuO + CO \to Cu + C{O_2}\\
2)\\
a)\\
N{a_2}C{O_3} + 2HCl \to 2NaCl + C{O_2} + {H_2}O\\
{n_{C{O_2}}} = \dfrac{{0,448}}{{22,4}} = 0,02\,mol\\
{n_{HCl}} = 2{n_{C{O_2}}} = 0,04\,mol\\
{C_M}HCl = \dfrac{{0,04}}{{0,02}} = 2M\\
b)\\
{n_{N{a_2}C{O_3}}} = {n_{C{O_2}}} = 0,02\,mol\\
\% {m_{N{a_2}C{O_3}}} = \dfrac{{0,02 \times 106}}{5} \times 100\% = 42,4\% \\
\% {m_{NaCl}} = 100 - 42,4 = 57,6\%
\end{array}\)
Em xem đề câu 2b lại nhé
