Đáp án:

 Bài 1: a) 40%

b) 0,25M

Bài 2: a) 18,96%

b) 4,08%

c) $C{\% _{NaOH}} = 3,08\% ;C{\% _{KOH}} = 1,5\% $

Bài 3: a) 0,5M

b) 6 g

Giải thích các bước giải:

 Bài 1:

a) ${n_{N{a_2}O}} = \dfrac{{3,1}}{{62}} = 0,05mol$

$N{a_2}O + {H_2}O \to 2NaOH$

$ \Rightarrow {n_{NaOH}} = 2{n_{N{a_2}O}} = 0,1mol$

${m_{dd}} = {m_{N{a_2}O}} + {m_{{H_2}O}} = 3,1 + 6,9 = 10g$

$ \Rightarrow C{\% _{NaOH}} = \dfrac{{0,1.40}}{{10}}.100\%  = 40\% $

b) ${n_{{H_2}S{O_4}}} = \dfrac{{4,9}}{{98}} = 0,05mol$

$ \Rightarrow {C_{M({H_2}S{O_4})}} = \dfrac{{0,05}}{{0,2}} = 0,25M$

Bài 2: 

a) 

$S = \dfrac{{{m_{CaC{l_2}}}}}{{{m_{{H_2}O}}}}.100 = 23,4 \Rightarrow {m_{CaC{l_2}}} = 0,234{m_{{H_2}O}}$

$\begin{gathered}
   \Rightarrow C\%  = \dfrac{{{m_{CaC{l_2}}}}}{{{m_{dd}}}}.100\%  = \dfrac{{{m_{CaC{l_2}}}}}{{{m_{CaC{l_2}}} + {m_{{H_2}O}}}}.100\%  \hfill \\
   = \dfrac{{0,234{m_{{H_2}O}}}}{{0,234{m_{{H_2}O}} + {m_{{H_2}O}}}}.100\%  = 18,96\%  \hfill \\ 
\end{gathered} $

b) Đặt ${n_{{H_2}S{O_4}}} = x$

$ \Rightarrow {V_{dd}} = \dfrac{x}{{0,5}}(l) = 2000x(ml) \Rightarrow {m_{dd}} = 2000x.1,2 = 2400x$

$ \Rightarrow C\%  = \dfrac{{98x}}{{2400x}}.100\%  = 4,08\% $

c) Đặt ${n_{KOH}} = x;{n_{NaOH}} = 2x$

$ \Rightarrow {V_{dd}} = \dfrac{{2x}}{1} = 2x(l) = 2000x(ml) \Rightarrow {m_{dd}} = 2000x.1,3 = 2600x$

$\begin{gathered}
   \Rightarrow C{\% _{NaOH}} = \dfrac{{40.2x}}{{2600x}}.100\%  = 3,08\%  \hfill \\
  C{\% _{KOH}} = \dfrac{{39x}}{{2600x}}.100\%  = 1,5\%  \hfill \\ 
\end{gathered} $

Bài 3: 

a) $S{O_3} + {H_2}O \to {H_2}S{O_4}$

$\begin{gathered}
  {n_{{H_2}S{O_4}}} = {n_{S{O_3}}} = \dfrac{{20}}{{80}} = 0,25mol \hfill \\
   \Rightarrow {C_M} = \dfrac{n}{V} = \dfrac{{0,25}}{{0,5}}0,5M \hfill \\ 
\end{gathered} $

b) $Mg + {H_2}S{O_4} \to MgS{O_4} + {H_2}$

$ \Rightarrow {n_{Mg}} = {n_{{H_2}S{O_4}}} = 0,25mol \Rightarrow {m_{Mg}} = 0,25.24 = 6g$