Đổi 200ml = 0,2 lít
Ta có: \(n_{CuSO_4}=0,5.0,2=0,1\left(mol\right)\)
PTHH: Fe + CuSO4 ---> FeSO4 + Cu
Theo PT: \(n_{Fe}=n_{CuSO_4}=0,1\left(mol\right)\)
\(\Rightarrow m_{Fe}=0,1.56=5,6\left(g\right)\)
Chọn B
\(n_{CuSO_4}=0.2\cdot0.5=0.1\left(mol\right)\)
\(Fe+CuSO_4\rightarrow FeSO_4+Cu\)
\(0.1.........0.1\)
\(m_{Fe}=0.1\cdot56=5.6\left(g\right)\)
\(B\)
\(Fe+CuSO_4\rightarrow FeSO_4+Cu\\ n_{Fe}=n_{CuSO_4}=0,5.0,2=0,1\left(mol\right)\\ m_{Fe}=0,1.56=5,6\left(g\right)\)
=> Chọn B
