`#Rick`

Theo đề bài ta có:

`n_{Na_{2}O}` `=` \(\dfrac{9,3}{62}=0,15\left(mol\right)\) ; \(n_{FeSO_{4}}=\dfrac{200.16}{152.100}\approx0,2\left(mol\right)\)

Ta có `\text{PTHH:}`

`(1)` \(Na_{2}O+H_{2}O\rightarrow2NaOH\left(\text{dd}A\right)\)

`\text{0,15mol.........................0,3mol}`

`1.`

Ta có:

\(C\%_{NaOH}=\dfrac{0,3.40}{9,3+90,7}.100\%=12\%\)

`2.`

Ta có `\text{PTHH 2:}`

\(2NaOH+FeSO4\rightarrow Na2SO4+Fe\left(OH\right)2\downarrow\)

Theo `\text{PTHH 2}` ta có: 

\(nNaOH=\dfrac{0,3}{2}mol< nFeSO4=\dfrac{0,2}{1}mol\)

`=> n_{FeSO_{4}` dư `(`tính theo `n_{NaOH}``)`

Ta có:

Kết tủa `B` thu được là `Fe(OH)_{2}` `=> m_{Fe(OH)_{2} ``= 0,15.90=13,5(g)`

`\text{DD C}` thu được là `FeSO_4` dư và `Na_2SO_4`

`=>` \(\left\{{}\begin{matrix}C\%FeSO_4\left(d\text{ư}\right)=\dfrac{\left(0,2-0,15\right).152}{0,3.40+200-13,5}.100\%\approx3,829\%\\C\%Na_2SO_4=\dfrac{0,15.142}{0,3.40+200-13,5}.100\%\approx10,73\%\end{matrix}\right.\)

`3.`

Ta có `\text{PTHH 3:}`

\(4Fe\left(OH\right)2+O2+2H2O-^{t0}\rightarrow4Fe\left(OH\right)3\)

`2Fe(OH)_3` $\to^{t^0}$ `Fe_2O_3+3H_2O`

Chấn rắn `D` là `Fe_2O_3`

Ta có:

`n_{Fe(OH)_2}` `=`` n_{Fe(OH)_{3} ``=`` 0,15 (mol)`

`n_{Fe_{2}O_{3}} =` \(\dfrac{1}{2}nFe\left(OH\right)3=\dfrac{1}{2}.0,15=0,075\left(mol\right)\)

`=> m_{Fe_{2}O_{3}} = 0,075.160=12(g)`

Đáp án:

Bạn tham khảo lời giải ở dưới nhé!!!

Giải thích các bước giải:

\(\begin{array}{l}
N{a_2}O + {H_2}O \to 2NaOH\\
2NaOH + FeS{O_4} \to N{a_2}S{O_4} + Fe{(OH)_2}\\
4Fe{(OH)_2} + {O_2} \to 2F{e_2}{O_3} + 4{H_2}O\\
{n_{N{a_2}O}} = 0,15mol\\
 \to {n_{NaOH}} = 2{n_{N{a_2}O}} = 0,3mol\\
 \to {m_{NaOH}} = 12g\\
 \to C{\% _{NaOH}} = \dfrac{{12}}{{9,3 + 90,7}} \times 100\%  = 12\% \\
{n_{FeS{O_4}}} = \dfrac{{200 \times 16}}{{100 \times 152}} = 0,21mol\\
 \to \dfrac{{{n_{NaOH}}}}{2} < {n_{FeS{O_4}}}
\end{array}\)

Suy ra \(FeS{O_4}\) dư

\(\begin{array}{l}
 \to {n_{Fe{{(OH)}_2}}} = \dfrac{1}{2}{n_{NaOH}} = 0,15mol\\
 \to {m_{Fe{{(OH)}_2}}} = 13,5g\\
 \to {n_{FeS{O_4}}}dư= 0,21 - \dfrac{1}{2}{n_{NaOH}} = 0,06mol\\
 \to {n_{N{a_2}S{O_4}}} = \dfrac{1}{2}{n_{NaOH}} = 0,15mol\\
{m_{{\rm{dd}}}} = 9,3 + 90,7 + 200 - 13,5 = 286,5g\\
 \to C{\% _{FeS{O_4}}}dư= \dfrac{{0,06 \times 152}}{{286,5}} \times 100\%  = 3,18\% \\
 \to C{\% _{N{a_2}S{O_4}}} = \dfrac{{0,15 \times 142}}{{286,5}} \times 100\%  = 7,43\% \\
F{e_2}{O_3} + 6HCl \to 2FeC{l_3} + 3{H_2}O\\
{n_{F{e_2}{O_3}}} = \dfrac{1}{2}{n_{Fe{{(OH)}_2}}} = 0,075mol\\
 \to {n_{HCl}} = 6{n_{F{e_2}{O_3}}} = 0,45mol\\
 \to {V_{HCl}} = \dfrac{{0,45}}{{1,5}} = 3l
\end{array}\)