Question

Giúp mình câu này với ạ, mình đang cần gấp ạ

Answer 1

a) 

Fe không phản ứng với H₂SO₄ đặc nguội

\(n_{SO_2}=\dfrac{1,176}{22,4}=0,0525\left(mol\right)\)

PTHH: Cu + 2H₂SO₄ --> CuSO₄ + SO₂ + 2H₂O

         0,0525<-0,105<----0,0525<-0,0525

\(\%m_{Cu}=\dfrac{0,0525.64}{27}.100\%=12,44\%\)

\(\%m_{Fe}=100\%-12,44\%=87,56\%\)

b) \(C_{M\left(dd.H_2SO_4\right)}=\dfrac{0,105}{0,8}=0,13125M\)

c) n_NaOH = 1,25.0,5 = 0,625 (mol)

PTHH: 2NaOH + CuSO₄ --> Cu(OH)₂ + Na₂SO₄

Xét tỉ lệ: \(\dfrac{0,625}{2}>\dfrac{0,0525}{1}\) => NaOH dư, CuSO₄ hết

PTHH: 2NaOH + CuSO₄ --> Cu(OH)₂ + Na₂SO₄

           0,105<---0,0525------------------>0,0525   

=> \(\left\{{}\begin{matrix}C_{M\left(NaOH_{dư}\right)}=\dfrac{0,625-0,105}{0,5}=1,04M\\C_{M\left(Na_2SO_4\right)}=\dfrac{0,0525}{0,5}=0,105M\end{matrix}\right.\)

 

Answer 2

a, n_H2 = \(\dfrac{\dfrac{1176}{1000}}{22,4}=0,0525\left(mol\right)\)

PTHH: Cu + 2H₂SO₄(đặc, nguội) ---> CuSO₄ + SO₂ + 2H₂O

       0,0525   0,105                              0,0525  0,0525

=> \(\left\{{}\begin{matrix}m_{Cu}=0,0525.64=3,36\left(g\right)\\m_{Fe}=27-3,36=23,64\left(g\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\%m_{Cu}=\dfrac{3,36}{27}=12,44\%\\\%m_{Fe}=100\%-12,44\%=87,56\%\end{matrix}\right.\)

b, \(C_{MddH_2SO_4}=\dfrac{0,105}{\dfrac{800}{1000}}0,13125M\)

c, n_NaOH = 1,25.\(\dfrac{500}{1000}\) = 0,625 (mol)

PTHH: CuSO₄ + 2NaOH ---> Cu(OH)₂ + Na₂SO₄

LTL: 0,0525 < \(\dfrac{0,625}{2}\) => NaOH dư

Theo pthh: \(\left\{{}\begin{matrix}n_{NaOH\left(pư\right)}=2n_{CuSO_4}=2.0,0525=0,105\left(mol\right)\\n_{Cu\left(OH\right)_2}=n_{Na_2SO_4}=n_{CuSO_4}=0,0525\left(mol\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}C_{MNaOH\left(dư\right)}=\dfrac{0,625-0,105}{0,5}=1,04M\\C_{MNa_2SO_4}=\dfrac{0,0525}{0,5}=0,105M\end{matrix}\right.\)