\(\begin{array}{l}
a)\\
160ml=0,16l\\
n_{Fe_2O_3}=\frac{16}{160}=0,1(mol)\\
n_{CuO}=\frac{6,4}{80}=0,08(mol)\\
n_{H_2SO_4}=0,16.2=0,32(mol)\\
Fe_2O_3+3H_2SO_4\to Fe_2(SO_4)_3+3H_2O\\
CuO+H_2SO_4\to CuSO_4+H_2O\\
\rm Nếu\,axit\,dư:\\
Theo\,PT:\,n_{H_2SO_4(pu)}=3n_{Fe_2O_3}+n_{CuO}=0,38>0,32\\
\to \rm Axit\,hết,\,oxit\,dư\\
TH_1:\,Fe_2O_3\,\rm pứ\,trước\\
Theo\,PT:\,n_{CuO(pu)}=0,32-3.0,1=0,02(mol)\\
\to n_{CuO(du)}=0,08-0,02=0,06(mol)\\
\to m=0,06.80=4,8(g)\\
TH_2:\,CuO\,\rm pứ\,trước\\
Theo\,PT:\,n_{Fe_2O_3(pu)}=\frac{0,32-0,08}{3}=0,08(mol)\\
\to n_{Fe_2O_3(du)}=0,08-0,06=0,02(mol)\\
\to m=0,02.160=3,2(g)\\
\to 3,2<m<4,8\\
b)\\
Dat\,V_{dd\,hh}=x(l)\\
\to n_{HCl}=x(mol);n_{H_2SO_4}=0,5x(mol)\\
\to n_H=x+0,5x.2=2x(mol)\\
2H+O\to H_2O\\
Theo\,PT:\,n_{O(oxit)}=\frac{1}{2}n_H=x(mol)\\
\to 3n_{Fe_2O_3}+n_{CuO}=x\\
\to x=0,08+0,1.3=0,38(l)
Vay\,can\,0,38l\,dd.
\end{array}\)