Giải thích các bước giải:

 Bài 4:

\(\begin{array}{l}
a)ZnO + {H_2} \to Zn + {H_2}O\\
{n_{ZnO}} = 0,15mol\\
 \to {n_{Zn}} = {n_{ZnO}} = 0,15mol\\
 \to {m_{Zn}} = 0,15 \times 65 = 9,75g\\
b)Zn + 2HCl \to ZnC{l_2} + {H_2}\\
{n_{HCl}} = 0,2mol\\
 \to {n_{HCl}} > {n_{Zn}} \to {n_{HCl}}dư\\
 \to {n_{{H_2}}} = {n_{Zn}} = 0,15mol\\
 \to {V_{{H_2}}} = 0,15 \times 22,4 = 3,36l
\end{array}\)

Bài 5:

\(\begin{array}{l}
2Na + 2{H_2}O \to 2NaOH + {H_2}\\
a)\\
{n_{Na}} = 0,1mol\\
 \to {n_{{H_2}}} = \dfrac{1}{2}{n_{Na}} = 0,05mol\\
 \to {V_{{H_2}}} = 0,05 \times 22,4 = 1,12l\\
b)\\
{n_{NaOH}} = {n_{Na}} = 0,1mol\\
 \to {m_{NaOH}} = 0,1 \times 40 = 4g\\
 \to {m_{{\rm{dd}}NaOH}} = {m_{Na}} + {m_{{H_2}O}} - {m_{{H_2}}} = 2,39 + 97,8 - 0,05 \times 2\\
 \to {m_{{\rm{dd}}NaOH}} = 100,09g\\
 \to C{\% _{{\rm{dd}}NaOH}} = \dfrac{4}{{100,09}} \times 100\%  = 4\% \\
\end{array}\)

Bài 3:

\(C\% KMn{O_4} = \dfrac{{40}}{{160 + 40}} \times 100\%  = 20\% \)