Question

Answer 1

a) 2Fe + 6H₂SO₄ --> Fe₂(SO₄)₃ + 3SO₂ + 6H₂O

2Fe0-6e-->Fe2+3	x1
S+6 +2e--> S+4	x3

 

b) \(n_{Fe}=\dfrac{14}{56}=0,25\left(mol\right)\)

\(n_{H_2SO_4}=0,032.2,5=0,08\left(mol\right)\)

PTHH: 2Fe + 6H₂SO₄ --> Fe₂(SO₄)₃ + 3SO₂ + 6H₂O

Xét tỉ lệ: \(\dfrac{0,25}{2}>\dfrac{0,08}{6}\) => Al dư, H₂SO₄ hết

PTHH: 2Fe + 6H₂SO₄ --> Fe₂(SO₄)₃ + 3SO₂ + 6H₂O

___________0,08------------------------->0,04

=> V_SO2 = 0,04.22,4 = 0,896(l)