Giải thích các bước giải:
\(\begin{array}{l}
2.\\
a)Ca + 2{H_2}O \to Ca{(OH)_2} + {H_2}\\
b)\\
{n_{Ca}} = 0,25mol\\
\to {n_{{H_2}}} = {n_{Ca}} = 0,25mol \to {V_{{H_2}}} = 5,6l\\
c)\\
{n_{Ca{{(OH)}_2}}} = {n_{Ca}} = 0,25mol\\
\to C{M_{Ca{{(OH)}_2}}} = \dfrac{{0,25}}{{0,5}} = 0,5M\\
d)Ca{(OH)_2} + {H_2}S{O_4} \to C{\rm{aS}}{O_4} + 2{H_2}O\\
{n_{{\rm{CaS}}{O_4}}} = {n_{Ca{{(OH)}_2}}} = 0,25mol\\
\to {m_{{\rm{CaS}}{O_4}}} = 34g\\
3.\\
a)Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
b)\\
{n_{{H_2}}} = 0,15mol\\
\to {n_{Fe}} = {n_{{H_2}}} = 0,15mol \to {m_{Fe}} = 8,4g\\
c)\\
{n_{FeS{O_4}}} = {n_{{H_2}}} = 0,15mol \to {m_{FeS{O_4}}} = 22,8g\\
d)CuO + {H_2} \to Cu + {H_2}O\\
{n_{CuO}} = {n_{{H_2}}} = 0,15mol \to {m_{CuO}} = 12g\\
\\
\end{array}\)
\(\begin{array}{l}
4.\\
a)Fe + 2HCl \to FeC{l_2} + {H_2}
\end{array}\)
Ag không tác dụng với dd HCl ở nhiệt độ thường
\(\begin{array}{l}
b)\\
{n_{{H_2}}} = 0,045mol\\
\to {n_{Fe}} = {n_{{H_2}}} = 0,045mol \to {m_{Fe}} = 2,52g\\
\to {m_{Ag}} = 4,68 - 2,52 = 2,16g \to {n_{Ag}} = 0,02mol\\
c)\\
Ag + 2HN{O_3} \to AgN{O_3} + N{O_2} + {H_2}O\\
{n_{N{O_2}}} = 2{n_{Ag}} = 0,04mol \to {V_{N{O_2}}} = 0,896l\\
d)\\
{n_{AgN{O_3}}} = {n_{Ag}} = 0,02mol \to {m_{AgN{O_3}}} = 3,4g
\end{array}\)
