\(4Al+3O_2\rightarrow\left(t^o\right)Al_2O_3\\ Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\\ a,n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\\ n_{O_2}=\dfrac{3}{4}.0,3=0,225\left(mol\right)\\ V_{O_2\left(\text{đ}ktc\right)}=0,225.22,4=5,04\left(l\right)\\ b,n_{Al_2O_3}=\dfrac{1}{4}.0,3=0,075\left(mol\right)\\ n_{H_2SO_4}=3.0,075=0,225\left(mol\right)\\ m_{H_2SO_4}=m=0,225.98=22,05\left(g\right)\)

Phương trình hóa học của phản ứng:

So sánh tỉ lệ  → Vậy Al2O3 dư

mAl2O3 (dư) = 60 - 17 = 43(g)

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

\(2:3:1:3\left(mol\right)\)

\(0,1:0,15:0,05:0,15\left(mol\right)\)

\(n_{H_2}=\dfrac{V}{22,4}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

\(a,m_{Al}=n.M=0,1.27=2,7\left(kg\right)\)

\(b,m_{Al_2\left(SO_4\right)_3}=n.M=0,05.342=17,1\left(g\right)\)

a)

b)

nAl=23nH2=2,4(mol)⇒mAl=2,4.27=64,8(gam)

c)

Theo PT trên : 

a)

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

b)

\(n_{H_2} = \dfrac{6.13,44}{22,4} = 3,6(mol)\)

Theo PTHH : 

\(n_{Al} = \dfrac{2}{3}n_{H_2} = 2,4(mol)\\ \Rightarrow m_{Al} = 2,4.27 = 64,8(gam)\)

c)

\(4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3\)

Theo PT trên : 

\(n_{O_2} = \dfrac{3}{4}n_{Al} = 1,8(mol)\\ \Rightarrow V_{O_2} = 1,8.22,4 = 40,32(lít)\)

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

Ta có: \(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{H_2SO_4}=0,15\left(mol\right)=n_{H_2}\\n_{Al_2\left(SO_4\right)_3}=0,05\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{H_2SO_4}=0,15\cdot98=14,7\left(g\right)\\m_{Al_2\left(SO_4\right)_3}=0,05\cdot342=17,1\left(g\right)\\V_{H_2}=0,15\cdot22,4=3,36\left(l\right)\end{matrix}\right.\)

a) \(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)

PTHH: 2Al + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂

______0,1--->0,15-------->0,05------->0,15

=> m_H2SO4 = 0,15.98 = 14,7 (g)

b) V_H2 = 0,15.22,4 = 3,36 (l)

c) m_Al2(SO4)3 = 0,05.342 = 17,1 (g)

`2Al + 3H_2 SO_4 -> Al_2 (SO_4)_3 + 3H_2`

`10 / 21`                                                        `5 / 7`      `(mol)`

`n_[H_2] = 16 / [ 22,4 ] = 5 / 7 (mol)`

`=> m_[Al] = 10 / 21 . 27 ~~ 12,9 (g)`

n H2=0,714 mol

2al+3h2so4 --> al2(so4)3 + 3h2

0,476------------------------------0,714

=>m Al=0,476.27=12,85g

\(n_{H_2}=\dfrac{16}{22,4}=\dfrac{5}{7}\left(mol\right)\)

Theo pthh: \(n_{Al}=\dfrac{2}{3}n_{H_2}=\dfrac{2}{3}.\dfrac{5}{7}=\dfrac{10}{21}\left(mol\right)\)

=> \(m_{Al}=\dfrac{10}{21}.27=\dfrac{90}{7}\left(g\right)\)

Ta có: \(n_{H_2}=\dfrac{74,37}{24,79}=3\left(mol\right)\)

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

a, \(n_{Al}=\dfrac{2}{3}n_{H_2}=2\left(mol\right)\)

\(\Rightarrow m_{Al}=2.27=54\left(g\right)\)

b, \(n_{H_2SO_4}=n_{H_2}=3\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4}=3.98=294\left(g\right)\)