Giải thích các bước giải:

Bài 4:

\(\begin{array}{l}
Mg + {H_2}S{O_4} \to MgS{O_4} + {H_2}\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
a)\\
{m_{{H_2}S{O_4}}} = \dfrac{{300 \times 9,8\% }}{{100\% }} = 29,4g\\
 \to {n_{{H_2}S{O_4}}} = 0,3mol\\
 \to {n_{{H_2}}} = {n_{{H_2}S{O_4}}} = 0,3mol\\
 \to {V_{{H_2}}} = 0,3 \times 22,4 = 6,72l\\
 \to {m_{{H_2}}} = 0,3 \times 2 = 0,6g\\
b)
\end{array}\)

Gọi a và b là số mol của Mg và Fe

\(\begin{array}{*{20}{l}}
{\left\{ {\begin{array}{*{20}{c}}
{24a + 56b = 10,4}\\
{a + b = 0,3}
\end{array}} \right. \to \left\{ {\begin{array}{*{20}{c}}
{a = 0,2}\\
{b = 0,1}
\end{array}} \right.}\\
{ \to {n_{Mg}} = 0,2mol \to {m_{Mg}} = 4,8g}\\
{{\rm{\;}} \to {n_{Fe}} = 0,1mol \to {m_{Fe}} = 5,6g}\\
{ \to \% {m_{Mg}} = \dfrac{{4,8}}{{10,4}} \times 100\% {\rm{ \;}} = 46,15\% }\\
{ \to \% {m_{Fe}} = \dfrac{{5,6}}{{10,4}} \times 100\% {\rm{ \;}} = 53,85\% }\\
{c)}\\
{{n_{MgS{O_4}}} = {n_{Mg}} = 0,2mol}\\
{{n_{FeS{O_4}}} = {n_{Fe}} = 0,1mol}\\
{ \to {m_{MgS{O_4}}} = 0,2 \times 120 = 24g}\\
{{\rm{\;}} \to {m_{FeS{O_4}}} = 0,1 \times 152 = 15,2g}\\
{ \to {m_{{\rm{dd}}}} = {m_{hỗnhợp}} + {m_{{\rm{dd}}{H_2}S{O_4}}} - {m_{{H_2}}} = 309,8g}\\
{ \to C{\% _{MgS{O_4}}} = \dfrac{{24}}{{309,8}} \times 100\% {\rm{ \;}} = 7,75\% }\\
{{\rm{\;}} \to C{\% _{FeS{O_4}}} = \dfrac{{15,2}}{{309,8}} \times 100\% {\rm{ \;}} = 4,91\% }
\end{array}\)

Bài 5:

\(\begin{array}{l}
Cu + 2{H_2}S{O_4} \to CuS{O_4} + 2{H_2}O + S{O_2}\\
CuO + {H_2}S{O_4} \to CuS{O_4} + {H_2}O\\
a)\\
{n_{S{O_2}}} = 0,2mol\\
 \to {n_{Cu}} = {n_{S{O_2}}} = 0,2mol\\
 \to {m_{Cu}} = 0,2 \times 64 = 12,8g\\
 \to {m_{CuO}} = 20,8 - 12,8 = 8g \to {n_{CuO}} = 0,1mol\\
b)\\
{n_{{H_2}S{O_4}}} = 2{n_{Cu}} + {n_{CuO}} = 0,5mol\\
 \to {m_{{H_2}S{O_4}}} = 0,5 \times 98 = 49g\\
 \to {m_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{{49}}{{80\% }} \times 100\%  = 61,25g\\
{n_{{\rm{CuS}}{{\rm{O}}_4}}} = {n_{Cu}} + {n_{CuO}} = 0,3mol\\
 \to {m_{{\rm{CuS}}{{\rm{O}}_4}}} = 0,3 \times 160 = 48g
\end{array}\)