\(n_{Zn}=\dfrac{9,75}{65}=0,15\left(mol\right)\\ pthh:Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)  
          0,15     0,15                        0,15 
\(V_{H_2}=0,15.22,4=3,36\left(l\right)\\ C_M=\dfrac{0,15}{0,1}=1,5M\)

Tham khảo: https://hoidap247.com/cau-hoi/1175785

\(a,PTHH:2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ \Rightarrow n_{Al}=\dfrac{2}{3}n_{H_2}=0,1\left(mol\right)\\ \Rightarrow m_{Al}=0,1\cdot27=2,7\left(g\right)\\ b,n_{H_2SO_4}=n_{H_2}=0,15\left(mol\right)\\ \Rightarrow m_{H_2SO_4}=0,15\cdot98=14,7\left(g\right)\\ c,n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{3}n_{H_2}=0,05\left(mol\right)\\ \Rightarrow m_{Al_2\left(SO_4\right)_3}=0,05\cdot342=17,1\left(g\right)\)

Gọi x,y lần lượt là số mol của Al, Fe

n_H2 = \(\dfrac{8,96}{22,4}\)=0,4 mol

Pt: 2Al + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂

......x.................................0,5x...........1,5x

.....Fe + H₂SO₄ --> FeSO₄ + H₂

.......y..........................y............y

Ta có hệ pt:

 {27x+56y=11

   1,5x+y=0,4

⇔x=0,2, y=0,1

% m_Al = \(\dfrac{0,2.27}{11}\).100%=49,1%

% m_Fe = \(\dfrac{0,1.56}{11}\).100%=50,9%

m_Al2(SO4)3 = 0,5x . 342 = 0,5 . 0,2 . 342 = 34,2 (g)

m_FeSO4 = 152y = 152 . 0,1 = 15,2 (g)

Gọi CTTQ: M_xO_y

Pt: M_xO_y + yH₂ --to--> xM + yH₂O

\(\dfrac{0,4}{y}\)<-------0,4

Ta có: 232,2=\(\dfrac{0,4}{y}\)(56x+16y)

⇔23,2=\(\dfrac{22,4x}{y}\)+6,4

⇔\(\dfrac{22,4x}{y}\)=16,8

⇔22,4x=16,8y

⇔x:y=3:4

Vậy CTHH của oxit: Fe₃O₄

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

PTHH: Zn + H₂SO₄ ---> ZnSO₄ + H₂

           0,1 <----------------------------- 0,1

=> m_Zn = 0,1.65 = 6,5 (g)

Zn+H2SO4->ZnSO4+H2

0,1----------------------------0,1

n H2=0,1 mol

=>m Zn=0,1.65=6,5g

\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ pthh:Zn+H_2SO_4->ZnCl_2+H_2\)
         0,1                                        0,1 
=>\(m_{Zn}=0,1.65=6,5 \left(g\right)\)

\(a.Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{Zn}=n_{ZnCl_2}=n_{H_2}=0,25\left(mol\right)\\ \Rightarrow m_{Zn}=0,25.65=16,25\left(g\right)\\ m_{ZnCl_2}=0,25.136=34\left(g\right)\\ b.FeO+H_2-^{t^o}\rightarrow Fe+H_2O\\ Tacó:n_{Fe}=n_{H_2}=0,25\left(mol\right)\\ \Rightarrow m_{Fe}=0,25.56=14\left(g\right)\)

\(n_{H_2SO_4}=0,2(mol)\\ \text{Bảo toàn H: }\\ n_{H_2}=n_{H_2SO_4}=0,2(mol)\\ V_{H_2}=0,2.22,4=4,48(l)\)

Fe + H\(_2\)SO\(_4\) → FeSO\(_4\) + H\(_2\)↑

Zn + H\(_2\)SO\(_4\) → ZnSO\(_4\) + H\(_2\)↑

nH\(_2\)SO\(_4\)= 0,1.2=0,2(mol)

Theo PT, nH\(_2\)SO\(_4\)= nH\(_2\)=0,2 (mol)

\(V\)\(H_2\)= 0,2 . 22,4= 4,48 (l)

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

\(PTHH:Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

(mol)_____0,2____0,2______0,2____0,2__

\(a.V_{H_2}=22,4.0,2=4,48\left(l\right)\)

\(b.m_{ddH_2SO_4}=\dfrac{0,2.98.100}{24,5}=80\left(g\right)\)

\(c.m_{ddspu}=13+80-0,2.2=92,6\left(g\right)\\ \Rightarrow C\%_{ddspu}=\dfrac{0,2.136}{92,6}.100=29,4\left(\%\right)\)