Nhận thấy $B+H_2SO_4$ ra kết tủa suy ra $B$ là $Ba(OH)_2$ dư

\(\begin{array}{l}
100ml=0,1l\\
Fe_2(SO_4)_3+3Ba(OH)_2\to 3BaSO_4↓+2Fe(OH)_3↓(1)\\
Ba(OH)_2+H_2SO_4\to BaSO_4↓+2H_2O(2)\\
2Fe(OH)_3\xrightarrow{t^o}Fe_2O_3+3H_2O(3)\\
n_{BaSO_4(2)}=\frac{9,32}{233}=0,04(mol)\\
Theo\,PT:\,n_{Ba(OH)_2(du)}=n_{BaSO_4(2)}=0,04(mol)\\
Dat\,n_{Fe_2(SO_4)_3}=x(mol);n_{Ba(OH)_2}=y(mol)\\
\to n_{Ba(OH)_2(pu)}=y-0,04(mol)\\
Theo\,PT:\,x=3(y-0,04)(1)\\
Theo\,PT:\,n_{BaSO_4(1)}=3x(mol);n_{Fe(OH)_3}=2x(mol)\\
\to n_{Fe_2O_3}=x(mol)\\
\to 3x.233+x.160=4,295\\
\to x=0,005(mol)\\
Thay\,vao\,(1)\,\to y=\frac{1}{24}(mol)\\
Theo\,PT:\,n_{H_2SO_4}=n_{BaSO_4(2)}=0,04(mol)\\
\to \begin{cases} C_{M\,Fe_2(SO_4)_3}=\frac{0,005}{0,1}=0,05M\\ C_{M\,Ba(OH)_2}=\frac{\frac{1}{24}}{0,1}=\frac{5}{12}M\\ V_{dd\,H_2SO_4}=\frac{0,04}{0,5}=0,08(l) \end{cases}
\end{array}\)