Em tham khảo nha :

\(\begin{array}{l}
a)\\
FeC{l_x} + xAgN{O_3} \to Fe{(N{O_3})_x} + xAgCl\\
b)\\
{n_{AgCl}} = \dfrac{{5,74}}{{143,5}} = 0,04mol\\
{n_{FeC{l_x}}} = \dfrac{{{n_{AgCl}}}}{x} = \frac{{0,04}}{x}mol\\
{M_{FeC{l_x}}} = 2,54:\dfrac{{0,04}}{x} = 63,5x\,dvC\\
 \Rightarrow 56 + 35,5x = 63,5x\\
 \Rightarrow x = 2\\
CTHH:FeC{l_2}\\
b)\\
{n_{AgN{O_3}}} = {n_{AgCl}} = 0,04mol\\
{m_{AgN{O_3}}} = 0,04 \times 170 = 6,8g\\
{m_{{\rm{dd}}AgN{O_3}}} = \dfrac{{6,8 \times 100}}{{25}} = 27,2g\\
d)\\
{n_{Fe{{(N{O_3})}_2}}} = \dfrac{{{n_{AgCl}}}}{2} = 0,02mol\\
{m_{Fe{{(N{O_3})}_2}}} = 0,02 \times 180 = 3,6g\\
C{\% _{Fe{{(N{O_3})}_2}}} = \dfrac{{3,6}}{{2,54 + 27,2 - 5,74}} \times 100\%  = 15\% 
\end{array}\)

Đáp án:

Giải thích các bước giải:

`a.`

`FeCl_x+xAgNO_3->xAgCl↓+Fe(NO_3)_x`

`b.`

`n_{AgCl}=(5,74)/(143,5)=0,04(mol)`

`n_{FeCl_x}=1/x n_{AgCl}=(0,04)/x(mol)`

`M_{FeCl_x}=(2,54)/((0,04)/x)=63,5x(g//mol)`

`=>56+35,5x=63,5x`

`=>x=2`

`=>FeCl_2`

`c.`

`n_{AgNO_3}=n_{AgCl}=0,04(mol)`

`m_{AgNO_3}=0,04.170=6,8(g)`

`m_{ddAgNO_3}=6,8:25%=27,2(g)`

`d.`

`m_{ddspư}=2,54+27,2-5,74=24(g)`

`n_{Fe(NO_3)_2}=0,5n_{AgCl}=0,02(mol)`

`C%_{ddspư}=(0,02.180)/24 .100=15%`