\(n_{Cu_2S}=n_{FeS}=a\left(mol\right)\\ 160a+88a=24,8\left(g\right)\\ \rightarrow n_{Cu_2S}=n_{FeS}=0,1\left(mol\right)\)

PTHH:

Cu₂S + 6H₂SO₄ ---> 2CuSO₄ + 5SO₂ + 6H₂O

0,1 ------> 0,6 ----------> 0,2 ------> 0,5 ------> 0,6

2FeS + 10H₂SO₄ ---> Fe₂(SO₄)₃ + 9SO₂ + 10H₂O

0,1 ------> 0,5 -----------> 0,05 --------> 0,45 ---> 0,5

\(\rightarrow\left\{{}\begin{matrix}V_{SO_2}=\left(0,5+0,45\right).22,4=21,28\left(l\right)\\n_{H_2SO_4}=0,5+0,6=1,1\left(mol\right)\end{matrix}\right.\)

PTHH:

Fe₂(SO₄)₃ + 6KOH ---> 2Fe(OH)₃ + 3K₂SO₄

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

CuSO₄ + 2KOH ---> Cu(OH)₂ + K₂SO₄

0,2 -----------------------> 0,2

2Fe(OH)₃ --t^o--> Fe₂O₃ + 3H₂O

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

Cu(OH)₂ --t^o--> CuO + H₂O

0,2 ----------------> 0,2

\(\rightarrow\left\{{}\begin{matrix}m=107.0,1+98.0,2=20,5\left(g\right)\\a=0,05.160+0,2.80=24\left(g\right)\end{matrix}\right.\)

\(m_{dd.H_2SO_4}=51,5.1,84=94,76\left(g\right)\)

\(n_{SO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: 2Al + 6H₂SO₄ --> Al₂(SO₄)₃ + 3SO₂ + 6H₂O

            0,2<----0,6<--------------------0,3

=> m_Al = 0,2.27 = 5,4 (g)

=> \(n_{Fe_2O_3}=\dfrac{21,4-5,4}{160}=0,1\left(mol\right)\)

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

                0,1--->0,3

=> \(m_{H_2SO_4\left(lý.thuyết\right)}=\left(0,6+0,3\right).98=88,2\left(g\right)\)

=> \(m_{H_2SO_4\left(tt\right)}=88,2.105\%=92,61\left(g\right)\)

=> \(C\%=\dfrac{92,61}{94,76}.100\%=97,7\%\)

Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

PTHH:

\(Fe+H_2SO_4--->FeSO_4+H_2\uparrow\left(1\right)\)

\(Cu+H_2SO_4--\times-->\)

Theo PT₍₁₎: \(n_{Fe}=n_{H_2}=0,1\left(mol\right)\)

\(\Rightarrow m_{Fe}=m_{Cu}=0,1.56=5,6\left(g\right)\)

PTHH: \(Cu+2H_2SO_{4_{đặc}}\overset{t^o}{--->}CuSO_4+SO_2\uparrow+2H_2O\left(2\right)\)

Ta có: \(n_{Cu}=\dfrac{5,6}{64}=0,0875\left(mol\right)\)

Theo PT₍₂₎: \(n_{SO_2}=n_{Cu}=0,0875\left(mol\right)\)

\(\Rightarrow V_{SO_2}=0,0875.22,4=1,96\left(lít\right)\)

a)

n KMnO4 = 0,26(mol)

Bảo toàn electron : 

5n KMnO4 = 2n SO2

<=> n SO2 = 0,26.5/2 = 0,65(mol)

V SO2 = 0,65.22,4 = 14,56 lít

b)

n H2SO4 pư = 2n SO2 = 0,65.2 = 1,3(mol)

n H2SO4 dư = 1,3.10% = 0,13(mol)

=> n H2SO4 đã dùng = 1,3 + 0,13 = 1,43(mol)

C% H2SO4 = 1,43.98/150   .100% = 93,43%

\(n_{SO_2}=\dfrac{V_{SO_2}}{22,4}=\dfrac{2,24}{22,4}=0,1mol\)

Gọi \(\left\{{}\begin{matrix}n_{Mg}=x\\n_{Cu}=y\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Mg}=24x\\m_{Cu}=64y\end{matrix}\right.\)

\(Mg+2H_2SO_4\rightarrow MgSO_4+SO_2+2H_2O\) 

 x           2x                               x                  ( mol )

\(Cu+2H_2SO_4\rightarrow CuSO_4+SO_2+2H_2O\)

 y            2x                              y                   ( mol )

Ta có:

\(\left\{{}\begin{matrix}24x+64y=4,4\\x+y=0,1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,05\\y=0,05\end{matrix}\right.\)

\(\Rightarrow m_{Mg}=0,05.24=1,2g\)

\(\Rightarrow m_{Cu}=0,05.64=3,2\)

\(m_{H_2SO_4}=n_{H_2SO_4}.M_{H_2SO_4}=\left(2.0,05+2.0,05\right).98=0,2.98=19,6g\)