Đáp án:

$V=291,67\\ m=36,04$

Giải thích các bước giải:

$Đặt\ n_{Fe_2O_3}=x(mol)\\ \Rightarrow n_{CuO}=y(mol)\\ \Rightarrow 160x+80.4x=20\\ \Rightarrow x=\dfrac{1}{24}\\ n_{HCl}=\dfrac{V}{1000}.1=0,001V(mol)\\ n_{H_2SO_4}=\dfrac{V}{1000}.0,5=0,0005V(mol)\\ \xrightarrow{\text{BTNT H}}\ n_{H^+}=n_{HCl}+2.n_{H_2SO_4}=0,002V(mol)\\ Fe_2O_3+6H^+\to 2Fe^{3+}+3H_2O\\ CuO+2H^+\to Cu^{2+}+H_2O\\ n_{H^+}=6.n_{Fe_2O_3}+2.n_{CuO}\\ =\dfrac{1}{24}.6+\dfrac{1}{24}.4.2\\=\dfrac{7}{12}(mol)\\ \Rightarrow \dfrac{7}{12}=0,002V\\ \Rightarrow V=291,67\\ n_{Fe^{3+}}=2.n_{Fe}=\dfrac{1}{12}(mol)\\ n_{Cu^{2+}}=n_{CuO}=\dfrac{1}{6}(mol)\\ \Rightarrow m_{muối}=\dfrac{1}{12}.162,5+\dfrac{1}{6}.135=36,04(g)\\ \Rightarrow m=36,04$