\(n_{Fe_2O_3}=\dfrac{8}{160}=0,05\left(mol\right)\)

PTHH: Fe₂O₃ + 3H₂SO₄ --> Fe₂(SO₄)₃ + 3H₂O

______0,05------>0,15--------->0,05

=> m_H2SO4 = 0,15.98 = 14,7(g)

=> \(C\%\left(H_2SO_4\right)=\dfrac{14,7}{100}.100\%=14,7\%\)

\(C\%\left(Fe_2\left(SO_4\right)_3\right)=\dfrac{0,05.400}{8+100}.100\%=18,52\%\)

PTHH: Fe₂(SO₄)₃ + 6NaOH --> 2Fe(OH)₃\(\downarrow\) + 3Na₂SO₄

________0,05----------------------->0,1

=> mFe(OH)₃ = 0,1.107=10,7(g)

\(n_{H_2SO_4}=\dfrac{98.5\%}{98}=0,05\left(mol\right)\\ PTHH:CuO+H_2SO_4\rightarrow CuSO_4+H_2O\\ n_{CuSO_4}=n_{CuO}=n_{H_2SO_4}=0,05\left(mol\right)\\ a,m_{CuO}=0,05.80=4\left(g\right)\\ b,m_{CuSO_4}=0,05.160=8\left(g\right)\\ m_{ddCuSO_4}=98+4=102\left(g\right)\\ C\%_{ddCuSO_4}=\dfrac{8}{102}.100\approx7,843\%\)

\(n_{FeO}=\dfrac{10,8}{72}=0,15\left(mol\right)\)

\(n_{H_2SO_4}=\dfrac{100.20\%}{100\%}:98=\dfrac{10}{49}\left(mol\right)\)

\(FeO+H_2SO_4\rightarrow FeSO_4+H_2O\)

0,15----> 0,15 -----> 0,15----->0,15

Xét \(\dfrac{10}{49}:1>\dfrac{0,15}{1}\) => axit dư.

Sau khi phản ứng kết thúc, trong dung dịch có:

\(m_{H_2SO_4}=\left(\dfrac{10}{49}-0,15\right).98=5,3\left(g\right)\)

\(m_{FeSO_4}=0,15.152=22,8\left(g\right)\)

\(m_{H_2O}=0,15.18=2,7\left(g\right)\)

\(n_{FeO}=\dfrac{10,8}{72}=0,15\left(mol\right)\)

\(n_{H_2SO_4}=\dfrac{100.20\%}{97}=\dfrac{10}{49}\left(mol\right)\)

PTHH :

                 FeO + H2SO4 → FeSO4 + H2O 

trc p/u:       0,15    10/49   

p/u :           0,15     0,15          0,15      0,15 

sau :         0           0,054        0,15      0,15 

\(m_{H2O}=0,15.18=2,7\left(g\right)\)

\(m_{ddFeSO_4}=10,8+100-2,7-5,292=102,808\left(g\right)\)

\(m_{H2SO4\left(dư\right)}=0,054.98=5,292\left(g\right)\)

Giúp mình với mn ơi 😭

Ta có nFe=7,2/72=0,1(mol)

Feo+H2SO4=>FeSO4+H2

a, nFeSO4=nFeO=0,1(mol)

=>mFeSO4=0,1.152=15,2(g)

b, ta có nH2SO4=0,1(mol)

=>Cm(H2SO4)=0,1/0,25=0,4(M)

\(n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\)

a) Pt : \(CuO+H_2SO_4\rightarrow CuSO_4+H_2O|\)  

             1            1                1            1

           0,2          0,2             0,2

b) \(n_{H2SO4}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

⇒ \(m_{H2SO4}=0,2.98=19,6\left(g\right)\)

\(C_{ddH2SO4}=\dfrac{19,6.100}{200}=9,8\)⁰/₀

c) \(n_{CuSO4}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

⇒ \(m_{CuSO4}=0,2.160=32\left(g\right)\)

\(m_{ddspu}=16+200=216\left(g\right)\)

\(C_{CuSO4}=\dfrac{32.100}{216}=14,81\)⁰/₀

\(m_{ct}=\dfrac{20.98}{100}=19,6\left(g\right)\)

\(n_{H2SO4}=\dfrac{19,6}{98}=0,2\left(mol\right)\)

Pt : \(CuO+H_2SO_4\rightarrow CuSO_4+H_2O|\)

         1             1               1             1

        0,2          0,2           0,2

a) \(n_{CuO}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

⇒ \(m_{CuO}=0,2.80=16\left(g\right)\)

b) \(n_{CuSO4}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

⇒ \(m_{CuSO4}=0,2.160=32\left(g\right)\)

c) \(m_{ddspu}=16+98=114\left(g\right)\)

\(C_{CuSO4}=\dfrac{32.100}{114}=28,7\)⁰/₀

 Chúc bạn học tốt

\(n_{CuO}=\dfrac{1.6}{80}=0.02\left(mol\right)\)

\(n_{H_2SO_4}=\dfrac{100\cdot20\%}{98}=\dfrac{10}{49}\left(mol\right)\)

\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

\(TC:\dfrac{0.02}{1}< \dfrac{10}{49}\Rightarrow H_2SO_4dư\)

\(m_{dd}=1.6+100=101.6\left(g\right)\)

\(C\%_{CuSO_4}=\dfrac{0.02\cdot160}{101.6}\cdot100\%=3.15\%\)

\(C\%_{H_2SO_4\left(dư\right)}=\dfrac{\left(\dfrac{10}{49}-0.02\right)\cdot98}{101.6}\cdot100\%=17.7\%\)