nFe=19/56=0,35(mol)
mCuSO4=100.10%.1,12=11,2(g).=>nCuSO4=\(\frac{11,2}{64+32+16\cdot4}=0,07\left(mol\right)\)
Fe+CuSO4----> Cu+FeSO4
Có: 0,35>0,07=> Fe dư
Fe+CuSO4----> Cu+FeSO4
0,07...0,07....................0,07(mol)
\(C_{M\left(FeSO_4\right)}=\frac{0,07}{0,1}=0,7\left(M\right)\)
#Walker
Ta có
\(n_{Fe}=\frac{19,6}{56}=0,35\left(mol\right)\)
\(m_{ddCuSO_4}=1,12\times100=112\left(g\right)\Rightarrow m_{CuSO_4}=\frac{10\times112}{100}=11,2\left(g\right)\)
\(\Rightarrow n_{CuSO_4}=\frac{11,2}{64+32+16\times4}=0,07\left(mol\right)\)
\(Fe+CuSO_4\rightarrow FeSO_4+Cu\)
Theo bài ra ta có \(n_{Fe}>n_{CuSO_4}\Rightarrow\) Tính theo \(CuSO_4\)
Theo PTHH \(n_{FeSO_4}=n_{CuSO_4}=0,07\left(mol\right)\)
\(\Rightarrow C_M\left(FeSO_4\right)=\frac{0,07}{0,1}=0.7M\)
Fe +CuSO4----->FeSO4 +Cu
ta có
m\(_{CuSO4}=100.1,12=112\left(g\right)\)
m\(_{ct}=\frac{112.10}{100}=11,2\left(g\right)\)
n\(_{Fe}=\frac{19,6}{56}=\)0,35 mol
n\(_{ct}=\frac{11,2}{160}=0,07mol\)
=> Fe dư
=> Các chất sau pư gồm Fe dư, FeSO4 và Cu
Theo thhh
n\(_{Fe}=n_{CuSO4}=0,07mol\)
=>n\(_{Fe}dư=0,35-0,07=0,28mol\)
C\(_MFe=\frac{0,28}{0,1}=2,8M\)
n\(_{FeSO4}=n_{Cu}=n_{Fe}=0,07mol\)
C\(_{M\left(FeSO4\right)}=C_{M\left(Cu\right)}=\frac{0,07}{0,1}=0,7M\)
nFe = 0.35 mol
mddCuSO4 = 112 g
mCuSO4 = 11.2 g
nCuSO4 = 0.07
Fe + CuSO4 --> FeSO4 + Cu
Bđ:0.35__0.07
Pư: 0.07__0.07_____0.07
Kt: 0.28___0_______0.07
CM FeSO4 = 0.07/0.1 = 0.7M
