Fe2O3 + 6HCl -> 2FeCl3 + 3H2O (1)
CuO + 2HCl -> CuCl2 + H2o (2)
nHCl=1,4(mol)
Đặt nFe2O3=a
nCuO=b
Ta có:
\(\left\{{}\begin{matrix}160a+80b=40\\6a+2b=1,4\end{matrix}\right.\)
=>a=0,2;b=0,1
mFe2O3=160.0,2=32(g)
mCuO=40-32=8(g)
Từ 1:
nFeCl3=2nFe2O3=0,4(mol)
Từ 2:
nCuCl2=nCuO=0,1(mol)
CM dd FeCl3=\(\dfrac{0,4}{0,7}=\dfrac{4}{7}M\)
CM dd CuCl2=\(\dfrac{0,1}{0,7}=\dfrac{1}{7}M\)