Đáp án:

Bạn tham khảo lời giải ở dưới nhé!!!

Giải thích các bước giải:

1,

a,

\(\begin{array}{l}
2Na + \dfrac{1}{2}{O_2} \to N{a_2}O\\
N{a_2}O + {H_2}O \to 2NaOH\\
2NaOH + {H_2}S{O_4} \to N{a_2}S{O_4} + 2{H_2}O\\
N{a_2}S{O_4} + BaC{l_2} \to 2NaCl + BaS{O_4}
\end{array}\)

b,

\(\begin{array}{l}
CuS{O_4} + BaC{l_2} \to CuC{l_2} + BaS{O_4}\\
CuC{l_2} + 2AgN{O_3} \to Cu{(N{O_3})_2} + 2AgCl\\
Cu{(N{O_3})_2} + 2NaOH \to 2NaN{O_3} + Cu{(OH)_2}\\
Cu{(OH)_2} \to CuO + {H_2}O
\end{array}\)

c,

\(\begin{array}{l}
2Al{(OH)_3} \to A{l_2}{O_3} + 3{H_2}O\\
A{l_2}{O_3} + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}O\\
A{l_2}{(S{O_4})_3} + 3BaC{l_2} \to 3BaS{O_4} + 2AlC{l_3}\\
AlC{l_3} + 3AgN{O_3} \to Al{(N{O_3})_3} + 3AgCl
\end{array}\)

d,

\(\begin{array}{l}
ZnO + 2HCl \to ZnC{l_2} + {H_2}O\\
ZnC{l_2} + 2NaOH \to 2NaCl + Zn{(OH)_2}\\
Zn{(OH)_2} + {H_2}S{O_4} \to ZnS{O_4} + 2{H_2}O\\
ZnS{O_4} + Ba{(N{O_3})_2} \to Zn{(N{O_3})_2} + BaS{O_4}
\end{array}\)

3,

\(\begin{array}{l}
2NaOH + {H_2}S{O_4} \to N{a_2}S{O_4} + 2{H_2}O\\
{n_{{H_2}S{O_4}}} = \dfrac{{200 \times 4,9}}{{100 \times 98}} = 0,1mol\\
 \to {n_{NaOH}} = 2{n_{{H_2}S{O_4}}} = 0,2mol\\
 \to {V_{NaOH}} = \dfrac{{0,2}}{2} = 0,1l\\
{n_{N{a_2}S{O_4}}} = {n_{{H_2}S{O_4}}} = 0,1mol\\
 \to {m_{N{a_2}S{O_4}}} = 14,2g
\end{array}\)

4,

\(\begin{array}{l}
2KOH + {H_2}S{O_4} \to {K_2}S{O_4} + 2{H_2}O\\
{n_{{H_2}S{O_4}}} = \dfrac{{300 \times 2,45}}{{100 \times 98}} = 0,075mol\\
 \to {n_{KOH}} = 2{n_{{H_2}S{O_4}}} = 0,15mol\\
 \to {V_{KOH}} = \dfrac{{0,15}}{{1,5}} = 0,1l\\
{n_{{K_2}S{O_4}}} = {n_{{H_2}S{O_4}}} = 0,075mol\\
 \to {m_{{K_2}S{O_4}}} = 13,05g
\end{array}\)