Em tham khảo nha :
\(\begin{array}{l}
17)\\
C\\
HCl + AgN{O_3} \to AgCl + HN{O_3}\\
{n_{HCl}} = 0,2 \times 0,2 = 0,04mol\\
{n_{AgCl}} = {n_{HCl}} = 0,04mol\\
{m_{AgCl}} = 0,04 \times 143,5 = 5,74g\\
18)\\
D\\
Zn + {H_2}S{O_4} \to ZnS{O_4} + {H_2}\\
{n_{Zn}} = \dfrac{{26}}{{65}} = 0,4mol\\
{n_{{H_2}}} = {n_{Zn}} = 0,4mol\\
{V_{{H_2}}} = 0,4 \times 22,4 = 8,96l\\
19)\\
C\\
{n_{S{O_2}}} = \dfrac{{38,4}}{{64}} = 0,6mol\\
{n_{NaOH}} = 0,75 \times 1 = 0,75mol\\
T = \dfrac{{{n_{NaOH}}}}{{{n_{S{O_2}}}}} = \dfrac{{0,75}}{{0,6}} = 1,25\\
\Rightarrow\text{Phản ứng tạo ra 2 muối } N{a_2}S{O_3},NaHS{O_3}\\
2NaOH + S{O_2} \to N{a_2}S{O_3} + {H_2}O\\
NaOH + S{O_2} \to NaHS{O_3}\\
hh:N{a_2}S{O_3}(a\,mol),NaHS{O_3}(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,6\\
2a + b = 0,75
\end{array} \right.\\
\Rightarrow a = 0,15;b = 0,45\\
{m_{N{a_2}S{O_3}}} = 0,15 \times 126 = 18,9g\\
{m_{NaHS{O_3}}} = 0,45 \times 104 = 46,8g\\
m = 18,9 + 46,8 = 65,7g\\
20)\\
A\\
Cu + 2{H_2}S{O_4} \to CuS{O_4} + S{O_2} + 2{H_2}O\\
FeO + {H_2}S{O_4} \to FeS{O_4} + {H_2}O\\
{n_{S{O_2}}} = \dfrac{{4,48}}{{22,4}} = 0,2mol\\
{n_{Cu}} = {n_{S{O_2}}} = 0,2mol\\
{m_{Cu}} = 0,2 \times 64 = 12,8g
\end{array}\)
