Giải thích các bước giải:

Bài 3:

\(\begin{array}{l}
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
{m_{HCl}} = \dfrac{{365 \times 10\% }}{{100\% }} = 36,5g\\
 \to {n_{HCl}} = 1mol\\
 \to {n_{{H_2}}} = {n_{ZnC{l_2}}} = {n_{Zn}} = \dfrac{1}{2}{n_{HCl}} = 0,5mol\\
 \to {V_{{H_2}}} = 0,5 \times 22,4 = 11,2l\\
 \to {m_{ZnC{l_2}}} = 0,5 \times 97 = 48,5g\\
 \to {m_{Zn}} = 0,5 \times 65 = 32,5g\\
 \to {m_{{\rm{ddZnC}}{{\rm{l}}_2}}} = {m_{Zn}} + {m_{{\rm{dd}}HCl}} - {m_{{H_2}}} = 32,5 + 365 - 0,5 \times 2\\
 \to {m_{{\rm{ddZnC}}{{\rm{l}}_2}}} = 396,5g\\
 \to C{\% _{{\rm{dd}}ZnC{l_2}}} = \dfrac{{48,5}}{{396,5}} \times 100\%  = 12,23\% 
\end{array}\)

Bài 4:

\(\begin{array}{l}
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
{n_{{H_2}S{O_4}}} = 0,27mol\\
 \to {n_{Al}} = \dfrac{2}{3}{n_{{H_2}S{O_4}}} = 0,18mol\\
 \to {m_{Al}} = 0,18 \times 27 = 4,86g\\
 \to {n_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{1}{3}{n_{{H_2}S{O_4}}} = 0,09mol\\
 \to C{M_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{{0,09}}{{0,2}} = 0,45M
\end{array}\)

Bài 3 *

Gọi X là kim loại có hóa trị I

\(\begin{array}{l}
2X + 2{H_2}O \to 2XOH + {H_2}\\
{n_{{H_2}}} = 0,15mol\\
 \to {n_X} = 2{n_{{H_2}}} = 0,3mol\\
 \to {M_X} = \dfrac{{6,9}}{{0,3}} = 23g\\
 \to X = Natri(Na)\\
PT:2Na + 2{H_2}O \to 2NaOH + {H_2}\\
 \to {V_{{H_2}O}} = {V_{NaOH}} = 0,2l\\
 \to {n_{NaOH}} = 2{n_{{H_2}}} = 0,3mol\\
 \to C{M_{NaOH}} = \dfrac{{0,3}}{{0,2}} = 1,5M
\end{array}\)

Bài 4*

\(\begin{array}{l}
AO + 2HN{O_3} \to A{(N{O_3})_2} + {H_2}O\\
{n_{HN{O_3}}} = 0,1mol\\
 \to {n_{AO}} = \dfrac{1}{2}{n_{HN{O_3}}} = 0,05mol\\
 \to {M_{AO}} = \dfrac{2}{{0,05}} = 40g\\
 \to {M_{AO}} = A + 16 = 40 \to A = 24\\
 \to A = Mg
\end{array}\)

Bài 5*

\(\begin{array}{l}
2A + 2{H_2}O \to 2AOH + {H_2}\\
{n_{{H_2}}} = 0,125mol\\
 \to {n_A} = 2{n_{{H_2}}} = 0,25mol\\
 \to {m_A} = 0,25 \times {M_A} = 5,75 \to {M_A} = 23\\
 \to A = Natri(Na)\\
PT:2Na + 2{H_2}O \to 2NaOH + {H_2}
\end{array}\)