Em tham khảo nha :

\(\begin{array}{l}
5)\\
a)\\
{K_2}O + {H_2}O \to 2KOH\\
{n_{{K_2}O}} = \dfrac{{9,4}}{{94}} = 0,1mol\\
{n_{KOH}} = 2{n_{{K_2}O}} = 0,2mol\\
{m_{KOH}} = 0,2 \times 56 = 11,2g\\
C{\% _{KOH}} = \dfrac{{11,2}}{{9,4 + 90,6}} \times 100\%  = 11,2\% \\
b)\\
CuS{O_4} + 2KOH \to Cu{(OH)_2} + {K_2}S{O_4}\\
{n_{CuS{O_4}}} = \dfrac{{{n_{KOH}}}}{2} = 0,1mol\\
{m_{CuS{O_4}}} = 0,1 \times 160 = 16g\\
{m_{ddCuS{O_4}}} = \dfrac{{16 \times 100}}{{16}} = 100g\\
{n_{Cu{{(OH)}_2}}} = \dfrac{{{n_{KOH}}}}{2} = 0,1mol\\
{m_{Cu{{(OH)}_2}}} = 0,1 \times 98 = 9,8g\\
{n_{{K_2}S{O_4}}} = \dfrac{{{n_{KOH}}}}{2} = 0,1mol\\
{m_{{K_2}S{O_4}}} = 0,1 \times 174 = 17,4g\\
C{\% _{{K_2}S{O_4}}} = \dfrac{{17,4}}{{100 + 100 - 9,8}} \times 100\%  = 9,15\% \\
6)\\
a)\\
{H_2}O + S{O_3} \to {H_2}S{O_4}\\
{n_{S{O_3}}} = \dfrac{{16}}{{80}} = 0,2mol\\
{n_{{H_2}S{O_4}}} = {n_{S{O_3}}} = 0,2mol\\
{m_{{H_2}S{O_4}}} = 0,2 \times 98 = 19,6g\\
C{\% _{{H_2}S{O_4}}} = \dfrac{{19,6}}{{16 + 184}} \times 100\%  = 9,8\% \\
b)\\
A + {H_2}S{O_4} \to AS{O_4} + {H_2}\\
{n_A} = {n_{{H_2}S{O_4}}} = 0,2mol\\
{M_A} = \dfrac{{4,8}}{{0,2}} = 24dvC\\
 \Rightarrow A:Magie(Mg)\\
c)\\
{n_{MgS{O_4}}} = {n_{{H_2}S{O_4}}} = 0,2mol\\
{m_{MgS{O_4}}} = 0,2 \times 120 = 24g\\
{m_{ddspu}} = 200 + 4,8 - 0,2 \times 2 = 204,4g\\
MgS{O_4} + 2NaOH \to Mg{(OH)_2} + N{a_2}S{O_4}\\
{m_{NaOH}} = \dfrac{{350 \times 12}}{{100}} = 42g\\
{n_{NaOH}} = \dfrac{{42}}{{40}} = 1,05mol\\
\dfrac{{0,2}}{1} < \dfrac{{1,05}}{2} \Rightarrow NaOH\text{ dư}\\
{n_{Mg{{(OH)}_2}}} = {n_{MgS{O_4}}} = 0,2mol\\
{m_{Mg{{(OH)}_2}}} = 0,2 \times 58 = 11,6g\\
{n_{NaO{H_d}}} = 1,05 - 0,2 \times 2 = 0,65mol\\
{m_{NaO{H_d}}} = 0,65 \times 40 = 26g\\
{n_{N{a_2}S{O_4}}} = {n_{MgS{O_4}}} = 0,2mol\\
{m_{N{a_2}S{O_4}}} = 0,2 \times 142 = 28,4\% \\
C{\% _{NaOH}} = \dfrac{{26}}{{204,4 + 350 - 11,6}} \times 100\%  = 4,8\% \\
C{\% _{N{a_2}S{O_4}}} = \dfrac{{28,4}}{{204,4 + 350 - 11,6}} \times 100\%  = 5,232\% 
\end{array}\)

Đáp án:

Giải thích các bước giải:

Câu 5:

`n_{K_2O}=\frac{9,4}{94}=0,1(mol)`

`K_2O+H_2O→2KOH`

`0,1:0,1:0,2`

`m_A=9,4+90,6=100(g)`

$a)$`C%_A=\frac{0,2.56}{100}.100=11,2%`

$b)$`CuSO_4+2KOH→K_2SO_4+Cu(OH)_2↓`

`0,1:0,2:0,1:0,1`

`m_{Cu(OH)_2}=0,1.98=9,8(g)`

`m_D=100+(0,1.160):16%-9,8=190,2(g)`

`C%_D=\frac{0,1.174}{190,2}.100=9,148%`

Câu 6:

`n_{SO_3}=\frac{16}{80}=0,2(mol)`

`SO_3+H_2O→H_2SO_4`

`0,2:0,2:0,2`

$a)$`C%_X=\frac{0,2.98}{16+184}.100=9,8%`

$b)$`A+H_2SO_4→ASO_4+H_2↑`

`0,2:0,2:0,2:0,2`

`M_A=\frac{4,8}{0,2}=24(g`$/$`mol)`

Vậy A là `Cu`

$c)$`n_{NaOH}=\frac{350.12%}{40}=1,05(mol)`

`2NaOH+MgSO_4→Mg(OH)_2↓+Na_2SO_4`

`\frac{1,05}2<\frac{0,2}1⇒NaOH\ \text{dư},\ MgSO_4\ \text{hết}`

`0,4:0,2:0,2:0,2`

`m_{Mg(OH)_2}=0,2.58=11,6(g)`

`m_D=16+184+4,8+350-0,2.2-11,6=542,8(g)`

`C%_{NaOH}=\frac{(1,05-0,4).40}{542,8}.100=4,79%`

`C%_{Na_2SO_4}=\frac{0,2.142}{542,8}.100=5,232%`

Xin hay nhất!!!