a,
Ta có :
\(\text{nFe2O3=16/16=0,1(mol)}\)
\(\text{nCuO=6,4/80=0,08(mol)}\)
\(\Rightarrow\)m muối=mFe2(SO4)3+mCuSO4=0,1x400+0,08x160=52,8(g)
b)
nH2SO4=0,1x3+0,08=0,38(mol)
\(\Rightarrow\text{CM=0,38/0,16=2,375(M)}\)
c)
Gọi a là V dd X
\(\Rightarrow\text{V+V=0,38=>V=0,19(l)}\)