bạn từng câu lên sẽ dễ nhìn hơn 

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

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

\(n_{Al}=\dfrac{16,2}{27}=0,6\left(mol\right)\)

2Al + 3H₂SO₄ ---> Al₂(SO₄)₃ + 3H₂

0,6       0,9                0,3            0,9

\(\rightarrow V_{H_2}=0,9.22,4=20,16\left(l\right)\)

\(n_{Cu}=\dfrac{57}{64}=0,890625\left(mol\right)\)

PTHH: CuO + H₂ --to--> Cu + H₂O

                0,890625          0,890625

\(H=\dfrac{0,890625}{0,9}=99\%\)

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 : 

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

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

$b) n_{Al} = \dfrac{10,8}{27} = 0,4(mol)$

Theo PTHH : $n_{H_2SO_4} = \dfrac{3}{2}n_{Al} = 0,6(mol)$
$m_{H_2SO_4} = 0,6.98 = 58,8(gam)$

$c) n_{Al_2(SO_4)_3} = \dfrac{1}{2}n_{Al} = 0,2(mol) \Rightarrow m_{Al_2(SO_4)_3} = 0,2.342 = 68,4(gam)$

$d) n_{H_2} = n_{H_2SO_4} = 0,6(mol) \Rightarrow V_{H_2} = 0,6.22,4 = 13,44(lít)$

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

\(b.n_{Al}=\dfrac{m}{M}=0,4\left(mol\right)\)

\(Theo.PTHH\Rightarrow n_{H_2SO_4}=n_{H_2}=\dfrac{3}{2}n_{Al}=1,5.0,4=0,6\left(mol\right)\\ \Rightarrow m_{H_2SO_4}=n.M=0,6.98=58,8\left(g\right)\)

\(c,Theo.PTHH\Rightarrow n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Al}=0,5.0,4=0,2\left(mol\right)\\ \Rightarrow m_{Al_2\left(SO_4\right)_3}=n.M=0,2.342=68,4\left(g\right)\\ d,V_{H_2\left(dktc\right)}=n.22,4=0,6.22,4=13,44\left(l\right)\)

Theo định luật bảo toàn khối lượng

⇒ \(m_{Al}+m_{H_2SO_4}=m_{Al_2\left(SO_4\right)_3}+m_{H_2}\)

⇒ \(m_{H_2}=0,3\left(g\right)\)

\(n_{H_2}=\dfrac{m}{M}=\dfrac{0,3}{2}=0,15\left(mol\right)\)

\(V_{H_2\left(đktc\right)}=n.22,4=0,15.22,4=3,36\left(l\right)\)

số phân tử khí H₂ trong phản ứng trên

\(=0,15.6.10^{23}=9.10^{22}\)(phân tử)

Theo ĐLBTKL: m_Al + m_H2SO4 = m_Al2(SO4)3 + m_H2

=> m_H2 = 2,7 + 14,7 - 17,1 = 0,3(g)

\(n_{H_2}=\dfrac{0,3}{2}=0,15\left(mol\right)\)

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

Số phân tử H₂ = 0,15.6.10²³ = 0,9.10²³

Đề sai nha bạn

nAl = 5.4/27 = 0.2 mol 

2Al + 3H2SO4 => Al2(SO4)3 + 3H2 

0.2______________0.1________0.3 

VH2 = 0.3*22.4 = 6.72 (l) 

mAl2(SO4)3 = 0.1*342 = 34.2 g