a) Fe + CuSO4 → FeSO4 + Cu
\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
\(n_{CuSO_4}=0,2\times1=0,2\left(mol\right)\)
Theo PT: \(n_{Fe}=n_{CuSO_4}\)
Theo bài: \(n_{Fe}=\dfrac{1}{2}n_{CuSO_4}\)
Vì \(\dfrac{1}{2}< 1\) ⇒ CuSO4 dư, Fe hết ⇒ Tính theo Fe
b) Theo PT: \(n_{Cu}=n_{Fe}=0,1\left(mol\right)\)
\(\Rightarrow m_{Cu}=0,1\times64=6,4\left(g\right)\)
Theo bài: nFe bđ = 5,6/56 = 0,1 mol ; nCuSO4 bđ = 0,2 .1 = 0,2mol
Fe + CuSO4 -> FeSO4 + Cu
trước pư: 1.........1................1............1 (mol)
tgpư: 0,1.............0,1............0,1........0,1 (mol)
dư: 0.................0,1...........0.................0 (mol)
Ta có:
nCu = nFe = 0,1 mol
=>mCu = 0,1 . 64 = 6,4 g
