Em tham khảo nha :

\(\begin{array}{l}
a)\\
N{a_2}O + {H_2}O \to 2NaOH\\
{n_{N{a_2}O}} = \dfrac{{15,5}}{{62}} = 0,25mol\\
{n_{NaOH}} = 2{n_{N{a_2}O}} = 0,5mol\\
{C_{{M_{NaOH}}}} = \dfrac{{0,5}}{{0,5}} = 1M\\
b)\\
2NaOH + {H_2}S{O_4} \to N{a_2}S{O_4} + 2{H_2}O\\
{n_{{H_2}S{O_4}}} = \dfrac{{{n_{NaOH}}}}{2} = 0,25mol\\
{m_{{H_2}S{O_4}}} = 0,25 \times 98 = 24,5g\\
{m_{dd{H_2}S{O_4}}} = \dfrac{{24,5 \times 100}}{{20}} = 122,5g\\
{V_{{H_2}S{O_4}}} = \dfrac{{122,5}}{{1,14}} = 107ml\\
c)\\
{n_{N{a_2}S{O_4}}} = \dfrac{{{n_{NaOH}}}}{2} = 0,25mol\\
{C_{{M_{N{a_2}S{O_4}}}}} = \dfrac{{0,25}}{{0,5 + 0,107}} = 0,412M
\end{array}\)

$\text{@Kenvin2007}$

Đáp án:

Câu 8 : 

a)

`Na_2``O` + `H_2``O` → `2``NaOH` 

`n_{Na_2O}` = `(15,5)/(62)` = `0,25` `mol` 

`n_(NaOH)` = `(0,25 × 2)/1` = `0,5` `mol` 

`C_{M_(NaOH)}` = `(0,5)/(0,5)` = `1` `M` 

b) 

`2``NaOH` + $H_2SO_4$ → $Na_2SO_4$ + `2``H_2``O`

`n_{H_2SO_4}` = `(0,5 × 1)/2` = `0,25` `mol` 

`m_{H_2SO_4}` = `0,25` × `98` = `24,5` `gam` 

`m_{dd_(H_2SO_4)}` = `(24,5 × 100)/(20)` = `122,5` `gam` 

`V_{H_2SO_4}` = `(122,5)/(1,14)` = `107` `ml` 

$\text{Đổi 107 ml = 0,107 l}$

c) 

`n_{Na_2SO_4}` = `(0,5 × 1)/2` = `0,25` `mol` 

`V_{Na_2SO_4}` = `0,5` + `0,107` = `0,607` `l` 

`C_{M_(Na_2SO_4)}` = `(0,25)/(0,607)` = `0,4118` `M` 

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