\(\left(a\right)2Al+3H_2O\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ \left(b\right)n_{H_2}=\dfrac{13,44}{22,4}=0,6mol\\ n_{Al}=\dfrac{0,6.2}{3}=0,4mol\\ m_{Al}=0,4.27=10,8g\\ \left(c\right)n_{O_2}=\dfrac{4,8}{32}=0,15mol\\ 4Al+3O_2\underrightarrow{t^0}2Al_2O_3\\ \Rightarrow\dfrac{0,4}{4}>\dfrac{0,15}{3}\Rightarrow Al.dư\\ n_{Al_2O_3}=\dfrac{0,15.2}{3}=0,1mol\\ m_{oxit}=m_{Al_2O_3}=0,1.102=10,2g\)

a: \(2Al+3H_2SO_4\rightarrow1Al_2\left(SO_4\right)_3+3H_2\uparrow\)

    0,4           0,6                0,2             0,6

b: \(n_{H_2}=\dfrac{13.44}{22.4}=0.6\left(mol\right)\)

=>\(n_{Al}=0.4\left(mol\right)\)

\(m_{Al}=0.4\cdot27=10.8\left(g\right)\)

c: \(4Al+3O_2\rightarrow2Al_2O_3\)

0,4                      0,2

\(m_{Al_2O_3}=0.2\left(27\cdot2+16\cdot3\right)=0.2\cdot102=20.4\left(g\right)\)

\(n_{H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

PTHH ;

2Al + 3H2SO4 --> Al2(SO4)3 + 3H2↑

  0,4           0,6               0,2           0,6

4Al + 3O2 ---> 2Al2O3

0,2       0,15          0,1 

\(n_{O_2}=\dfrac{4,8}{32}=0,15\left(mol\right)\)

\(\dfrac{0,4}{4}>\dfrac{0,15}{3}\)

--> Tính theo oxi

\(b,m_{Al}=0,4.27=10,8\left(g\right)\)

\(c,m_{Al_2O_3}=0,1.102=10,2\left(g\right)\)

a)

$n_{Al} = 0,3(mol)$

$2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2$

Theo PTHH : 

$n_{H_2SO_4} = \dfrac{3}{2}n_{Al} = 0,45(mol)$
$m_{dd\ H_2SO_4} = \dfrac{0,45.98}{12,25\%} = 360(gam)$

b)

$n_{H_2} = n_{H_2SO_4} = 0,45(mol)$
$V_{H_2} = 0,45.22,4 = 10,08(lít)$

c)

$n_{Al_2(SO_4)_3} = 0,15(mol)$
$m_{dd\ sau\ pư} = 8,1 + 360 - 0,45.2 = 367,2(gam)$
$C\%_{Al_2(SO_4)_3} = \dfrac{0,15.342}{367,2}.100\% = 14\%$

a)   2Al +  6HCl   →  2AlCl₃  + 3H₂ 

b)    nHCl = \(\dfrac{65,7}{36,5}\)= 1,8 mol

Theo tỉ lệ phản ứng => nAl phản ứng = \(\dfrac{nHCl}{3}\)= 0,6 mol

=> mAl phản ứng = 0,6.27 = 16,2 gam

c) nH₂ = 1/2nHCl = 0,9 mol

=> V H₂ = 0,9.22,4 = 20,16 lít

a.b.c.\(n_{Al}=\dfrac{2,7}{27}=0,1mol\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

0,1                         0,1      0,15   ( mol )

\(V_{H_2}=0,15.22,4=3,36l\)

\(m_{AlCl_3}=0,1.133,5=13,35g\)

d.\(Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\)

                  0,15              0,1              ( mol )

\(m_{Fe}=0,1.56=5,6g\)

2Al+6HCl-->2AlCl₃+3H₂

0,3----0,9---------0,3------0,45

=>n _Al=8,1\17=0,3 mol

=>V_H2=0,45.22,4=10,08l

=>m _HCl=0,9.26,5=32,85g

=>m_AlCl3=0,3.133,5=40,05g

C2 :Bảo Toàn khối lượng 

=>m _AlCl3=40,05g

B nha

a. 2Al + 6HCl -> 2AlCl₃ + 3H₂

b. n_Al = \(\dfrac{8.1}{27}=0,3\left(mol\right)\)=> \(n_{H_2}=\dfrac{3}{2}.0,3=0,45\left(mol\right)\)

\(V_{H_2}=0,45.22,4=10,08\left(mol\right)\)

c. \(n_{HCl}=3n_{Al}=3.0,3=0,9\left(mol\right)=>m_{HCl}=0,9.36,5=32,85\left(g\right)\)

Vậy m = 32,85

Sửa đề thành 0,54 gam Al cho số mol đẹp bạn nhé!

PT: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

Ta có: \(n_{Al}=\dfrac{0,54}{27}=0,02\left(mol\right)\)

Theo PT: \(\left\{{}\begin{matrix}n_{H_2SO_4}=n_{H_2}=\dfrac{3}{2}n_{Al}=0,03\left(mol\right)\\n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Al}=0,01\left(mol\right)\end{matrix}\right.\)

a, Ta có: \(n_{H_2}=0,03.22,4=0,672\left(l\right)\)

b, \(m_{H_2SO_4}=0,03.98=2,94\left(g\right)\)

c, \(m_{Al_2\left(SO_4\right)_3}=0,01.342=3,42\left(g\right)\)

Bạn tham khảo nhé!

1.

a, \(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\)

PTHH: 2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂

Mol:      0,3                          0,15          0,45

b, \(V_{H_2}=0,45.22,4=10,08\left(l\right)\)

c, Al₂(SO₄)₃ : nhôm sunfat

\(m_{Al_2\left(SO_4\right)_3}=0,15.342=51,3\left(g\right)\)

3.

a, \(n_{Cu}=\dfrac{38,4}{64}=0,6\left(mol\right)\)

PTHH: 2Cu + O₂ ---t^o→ 2CuO

Mol:      0,6     0,3        

CuO: đồng(ll) oxit

b, \(V_{O_2}=0,3.22,4=6,72\left(l\right)\)

c, 

PTHH: 2KMnO₄ ---t^o→ K₂MnO₄ + MnO₂ + O₂

Mol:        0,6                                                 0,3

\(m_{KMnO_4}=0,6.158=47,4\left(g\right)\)

2.

a, \(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\)

PTHH: Zn + 2HCl → ZnCl₂ + H₂

Mol:     0,3                   0,3       0,3

b, \(m_{ZnCl_2}=0,3.136=40,8\left(g\right)\)

c, 

PTHH: Fe₂O₃ + 3H₂ ---t^o→ 2Fe + 3H₂O

Mol:      0,1         0,3

\(m_{Fe_2O_3}=0,1.160=16\left(g\right)\)