Đáp án:

 Bạn tham khảo lời giải ở dưới nhé

Giải thích các bước giải:

1,

a,

Ta có: Khối lượng riêng của nước: \(D = 1g/ml\)

Suy ra:

\(\begin{array}{l}
 \to {m_{{H_2}O}} = 200g\\
 \to C{\% _{NaCl}} = \dfrac{{11,7}}{{11,7 + 200}} \times 100\%  = 5,527\% 
\end{array}\)

b,

Ta có: Khối lượng riêng của nước: \(D = 1g/ml\)

Suy ra:

\(\begin{array}{l}
 \to {m_{{H_2}O}} = 100g\\
 \to C{\% _{NaCl}} = \dfrac{{2,4}}{{2,4 + 100}} \times 100\%  = 2,344\% 
\end{array}\)

\(\begin{array}{l}
2.\\
Na + {H_2}O \to NaOH + \frac{1}{2}{H_2}\\
a)\\
{n_{Na}} = 0,4mol\\
 \to {n_{{H_2}}} = \dfrac{1}{2}{n_{Na}} = 0,2mol\\
 \to {V_{{H_2}}} = 4,48l\\
b)\\
{n_{NaOH}} = {n_{Na}} = 0,4mol \to {m_{NaOH}} = 16g\\
 \to {m_{{\rm{dd}}}} = {m_{Na}} + {m_{{H_2}O}} - {m_{{H_2}}} = 48,8g\\
 \to C{\% _{NaOH}} = \dfrac{{16}}{{48,8}} \times 100\%  = 32,8\% 
\end{array}\)

\(\begin{array}{l}
3.\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
a)\\
{n_{Mg}} = 0,2mol\\
 \to {n_{HCl}} = 2{n_{Mg}} = 0,4mol \to {m_{HCl}} = 14,6g\\
 \to {m_{{\rm{dd}}HCl}} = \dfrac{{14,6}}{{20\% }} \times 100\%  = 73g\\
b)\\
{n_{MgC{l_2}}} = {n_{Mg}} = 0,2mol \to {m_{MgC{l_2}}} = 19g\\
{n_{{H_2}}} = {n_{Mg}} = 0,2mol\\
{m_{{\rm{dd}}}} = {m_{Mg}} + {m_{{\rm{dd}}HCl}} - {m_{{H_2}}} = 77,4g\\
 \to C{\% _{MgC{l_2}}} = \dfrac{{19}}{{77,4}} \times 100\%  = 24,55\% \\
4.\\
S{O_3} + {H_2}O \to {H_2}S{O_4}\\
{n_{S{O_3}}} = 0,25mol\\
a)\\
{n_{{H_2}S{O_4}}} = {n_{S{O_3}}} = 0,25mol\\
 \to C{M_{{H_2}S{O_4}}} = \dfrac{{0,25}}{{0,2}} = 1,25M\\
b)\\
F{e_2}{O_3} + 3{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3{H_2}O\\
\left\{ \begin{array}{l}
{V_{{H_2}S{O_4}}} = 0,2l \to {n_{{H_2}S{O_4}}} = 0,25mol\\
{V_{{H_2}S{O_4}}} = 0,1l \to {n_{{H_2}S{O_4}}} = 0,125mol
\end{array} \right.\\
 \to {n_{F{e_2}{O_3}}} = \dfrac{1}{3}{n_{{H_2}S{O_4}}} = 0,042mol\\
 \to {m_{F{e_2}{O_3}}} = 6,72g\\
{n_{F{e_2}{{(S{O_4})}_3}}} = \dfrac{1}{3}{n_{{H_2}S{O_4}}} = 0,042mol\\
 \to C{M_{F{e_2}{{(S{O_4})}_3}}} = \dfrac{{0,042}}{{0,1}} = 0,42M\\
\end{array}\)