\(n_{Fe.pu}=x\)
\(n_{Fe}=\frac{8}{56}=\frac{1}{7}\left(mol\right)\)
\(PTHH:Fe+CuSO_4\rightarrow Cu+FeSO_4\)
(mol) x x x x
Theo đề bài ta có:
\(64x-56x=8,8-8\Leftrightarrow x=\frac{0,8}{8}=0,1\left(mol\right)\)
\(m_{Cu}=64x=64.0,1=6,4\left(g\right)\\ m_{Fe.du}=\left(\frac{1}{7}-0,1\right).56=2,4\left(g\right)\)
\(V_{ddspu}=22,4.0,1+0,5-22,4.0,1=0,5\left(l\right)\)
\(C_{M_{FeSO_4}}=\frac{0,1}{0,5}=0,2\left(M\right)\)
\(C_{M_{CuSO_4}}=\frac{\left(2.0,5\right)-0,1}{0,5}=1,8\left(M\right)\)
nCuSO4 = 1 mol
Đặt :
nFe (pư) = x mol
Fe + CuSO4 --> FeSO4 + Cu
x_____x________x______x
m tăng = mCu - mFe = 8.8 - 8 = 0.8
<=> 64x - 56x = 0.8
<=> x = 0.1 mol
mCu = 0.1*64 = 6.4 g
CM CuSO4 dư = ( 1 - 0.1 ) / 0.5=1.8M
CM FeSO4 = 0.1/0.5=0.2M