\(Fe+CuSO_4\rightarrow FeSO_4+Cu\)
0,1____0,1_______0,1________
\(n_{Fe}=\frac{5,6}{56}=0,1\left(mol\right)\)
\(n_{CuSO4}=\frac{200.1,12.10\%}{160}=0,14\left(mol\right)\)
Nên Fe hết, CuSO4 dư
\(\Rightarrow n_{CuSO4\left(dư\right)}=0,14-0,1=0,04\left(mol\right)\)
\(CM_{FeSO4}=\frac{0,1}{0,2}=0,5\left(M\right)\)
\(CM_{CuSO4}=\frac{0,04}{0,2}=0,2\left(M\right)\)