nFe=5,6/56=0,1mol
mCuSO4=32.100/100=32g
nCuSO4=32/160=0.2mol
Ptpu
Fe+CuSO4->FeSO4+Cu
0,1......0,1.........0,1.....0,1(mol)
mFeSO4=0,1.152=15,2g
mdung dịch sau pu=mFe+mddCuSO4-mCu
=5,6+100-0,1.64=99,2g
C%FeSO4=0,1.152.100/99,2=15,32%
Fe + CuSO4 → FeSO4 + Cu
\(m_{CuSO_4}=100\times32\%=32\left(g\right)\)
\(\Rightarrow n_{CuSO_4}=\dfrac{32}{160}=0,2\left(mol\right)\)
\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
Theo PT: \(n_{Fe}=n_{CuSO_4}\)
Theo bài: \(n_{Fe}=\dfrac{1}{2}n_{CuSO_4}\)
Vì \(\dfrac{1}{2}< 1\) ⇒ CuSO4 dư
Theo PT: \(n_{Cu}=n_{Fe}=0,1\left(mol\right)\)
\(\Rightarrow m_{Cu}=0,1\times64=6,4\left(g\right)\)
Dung dịch sau phản ứng gồm: CuSO4 dư và FeSO4
Theo PT: \(n_{CuSO_4}pư=n_{Fe}=0,1\left(mol\right)\)
\(\Rightarrow n_{CuSO_4}dư=0,2-0,1=0,1\left(mol\right)\)
\(\Rightarrow m_{CuSO_4}dư=0,1\times160=16\left(g\right)\)
Theo PT: \(n_{FeSO_4}=n_{Fe}=0,1\left(mol\right)\)
\(\Rightarrow m_{FeSO_4}=0,1\times152=15,2\left(g\right)\)
\(\Rightarrow\Sigma m_{ct}=15,2+16=31,2\left(g\right)\)
\(\Sigma m_{dd}=5,6+100-6,4=99,2\left(g\right)\)
\(\Rightarrow C\%_{dd}=\dfrac{31,2}{99,2}\times100\%=31,45\%\)