\(\begin{array}{l}
FeCl_x+xNaOH\to Fe(OH)_x+xNaCl\\
Theo\,PT:\,n_{FeCl_x}=n_{Fe(OH)_x}\\
\to \dfrac{6,5}{56+35,5x}=\dfrac{4,28}{56+17x}\\
\to x=3\\
\to CTHH:\,FeCl_3
\end{array}\)

\(\begin{array}{l}
200ml=0,2l\\
n_{SO_3}=\frac{8}{80}=0,1(mol)\\
SO_3+H_2O\to H_2SO_4(1)\\
2NaOH+H_2SO_4\to Na_2SO_4+2H_2O\\
Theo\,PT:\,n_{H_2SO_4(1)}=n_{SO_3}=0,1(mol)\\
\to \sum n_{H_2SO_4}=0,2.3+0,1=0,7(mol)\\
Theo\,PT:\,n_{NaOH}=2\sum n_{H_2SO_4}=1,4(mol)\\
\to m_{dd\,NaOH}=\frac{1,4.40}{40\%}=140(g)
\end{array}\)

\(\begin{array}{l}
200ml=0,2l\\
n_{H_2SO_4(bđ)}=0,2.3=0,6(mol)\\
n_{SO_3}=\frac{8}{80}=0,1(mol)\\
SO_3+H_2O\to H_2SO_4(1)\\
Theo\,PT:\,n_{H_2SO_4(1)}=n_{SO_3}=0,1(mol)\\
\to C_{M\,C}=C_{M\,H_2SO_4(sau)}=\frac{0,1+0,6}{0,2}=3,5M
\end{array}\)

\(\begin{array}{l}
a)\\
m_{X_2O_5}=76,4-0,3.160=28,4(g)\\
\to M_{X_2O_5}=2M_X+16.5=\frac{28,4}{0,2}=142(g/mol)\\
\to M_X=31(P)\\
\to X:\,photpho\,(P)\\
b)\\
\%m_{Fe_2O_3}=\frac{0,3.160}{76,4}.100\%\approx 62,83\%\\
\%m_{P_2O_5}=100-62,83=37,17\%
\end{array}\)

ai buff v :0

bỏ nghiệm (1;26) đi nhá :V

\[\begin{array}{l}
\begin{cases} \left(x+2\right)^2-x\left(y+1\right)+y=8\left(1\right)\\ 4x^2-24x+35=5\left(\sqrt{3y-14}+\sqrt{y-1}\right)\left(2\right) \end{cases}\\
ĐK:\,y\ge \dfrac{14}{3}\\
\left(1\right)\leftrightarrow \left[\left(x+2\right)^2-9\right]+\left(y-xy-x+1\right)=0\\
\leftrightarrow \left(x-1\right)\left(x+5\right)-\left(x-1\right)\left(y+1\right)=0\\
\leftrightarrow \left(x-1\right)\left(x-y+4\right)=0\\
\to \left[ \begin{array}{l}x=1\\y=x+4\end{array} \right.\\
Voi\,x=1\,thay\,vao\,PT(2):\\
\left(2\right)\leftrightarrow 4-24+35=5\left(\sqrt{3y-14}+\sqrt{y-1}\right)\\ \leftrightarrow \sqrt{3y-14}=3-\sqrt{y-1}\\ \leftrightarrow 3y-14=y+8-6\sqrt{y-1}\\ \leftrightarrow 11-y=3\sqrt{y-1}\\
\leftrightarrow y^2-22y+121=9y-9\\
\leftrightarrow y^2-31y+130=0\\
\leftrightarrow \left(y-26\right)\left(y-5\right)=0\\
\to y=26;y=5\\
Voi\,y=x+4\,thay\,vao\,PT(2):\\
\leftrightarrow \left(5-5\sqrt{3x-2}\right)+\left(10-5\sqrt{x+3}\right)+4x^2-24x+20=0\\ 
\leftrightarrow 4\left(x-5\right)\left(x-1\right)-5\left[\dfrac{3\left(x-1\right)}{1+\sqrt{3x-2}}+\dfrac{x-1}{2+\sqrt{x+3}}\right]=0\\ 
\to \left[\begin{array}{l}x=1\\4\left(x-5\right)-5\left(\dfrac{3}{1+\sqrt{3x-2}}+\dfrac{1}{2+\sqrt{x+3}}\right)=0\left(3\right)\end{array}\right.\\ 
\left(3\right)\leftrightarrow 4x-20-\dfrac{15}{1+\sqrt{3x-2}}-\dfrac{5}{2+\sqrt{x+3}}=0\\ \leftrightarrow \left(3-\dfrac{15}{1+\sqrt{3x-2}}\right)+\left(1-\dfrac{5}{2+\sqrt{x+3}}\right)+4x-24=0\\ 
\leftrightarrow \dfrac{3\left(\sqrt{3x-2}-4\right)}{1+\sqrt{3x-2}}+\dfrac{\sqrt{x+3}-3}{2+\sqrt{x+3}}+4x-24=0\\
\leftrightarrow \dfrac{9\left(x-6\right)}{\left(1+\sqrt{3x-2}\right)\left(\sqrt{3x-2}+4\right)}+\dfrac{x-6}{\left(2+\sqrt{x+3}\right)\left(\sqrt{x+3}+3\right)}+4\left(x-6\right)=0\\
\to \left[\begin{array}{l}x=6\to y=10\\\dfrac{9}{\left(1+\sqrt{3x-2}\right)\left(\sqrt{3x-2}+4\right)}+\dfrac{1}{\left(2+\sqrt{x+3}\right)\left(\sqrt{x+3}+3\right)}+4=0\left(\text{vô nghiệm}\right)\end{array}\right.
\end{array}\]

Vậy \(\left(x;y\right)=\left\{\left(6;10\right);\left(1;26\right);\left(1;5\right)\right\}\)

Bán phản ứng:

\(2Na+2H_2O\to 2Na^++2OH^-+H_2\)
\(Ba+2H_2O\to Ba^{2+}+2OH^-+H_2\)