How to Balance Al + H2SO4 = Al2(SO4)3 +H2

When the Sulfuric acid is diluted, the Aluminium reacts with the diluted Sulfuric acid (Al H2SO4) to produce Aluminium sulfate as well as Hydrogen gas. The reaction is a Redox reaction. So The reaction is-

Al + H₂SO₄ (dil.) = Al₂(SO₄)₃ +  H₂↑

Click to see what occurs when focused sulfuric acid– Al + H₂SO₄ (conc.) =?

The above reaction is an Oxidation-Reduction reaction because-

1. The Al oxidation number in the reactants is zero, while Al on the left hand is + 3 in the aluminum sulfate.
2. On the other hand, the Oxidation number of Hydrogen in the reactants is +1 but on the left side, the Oxidation number is reduced to 0. That means there is a change of electron between the reactants. That’s why the reaction is a Redox reaction.

Metal aluminum and sulfuric acid diluted (Al H2SO4 dil.)

Aluminium Metal is soft,  silvery-white as well as nonmagnetic. It is a very useful metal. It is an element of the periodic table having atomic number 13.

On the other hand Diluted Sulfuric acid is a mineral acid. It is a colorless, odorless, and syrupy liquid that is soluble in water. The Diluted Sulfuric Acid acts only as an acid in this chemical reaction. Diluted Sulfuric acid does not have oxidizing power.

Like as other acids Diluted Sulfuric acid reacts with the aluminum and produce Hydrogen gas.

Balancing the equation of the reaction:

As the reaction is a redox reaction, the equation of the reaction can be balanced using the ion-electron method.

After balancing the equation of the reaction-

2Al + 3H₂SO₄ (dil.) = Al₂(SO₄)₃ + 3H₂↑

Balancing (Using ion-Electron Method)

we can simple prof the above balance of the reaction equation using the Ion-Electron Method-

The skeleton reaction of between Aluminium and diluted Sulfuric acid (Al H2SO4) as follow-

Al + H₂SO₄ (dil.) = Al₂(SO₄)₃ + H₂↑

Here,

The Oxidizing agent: H₂SO₄ or H⁺
The Reducing agent: Al

Oxidation Half Reaction: Each Al gives up three electrons and the oxidation state has become +3 

⇒ Al – 3e^– = Al³⁺ … … … … … (1)

Reduction Half Reaction: Each three H⁺ ion taken up the three electrons given up by an Al atom. Then, the H⁺ ions become Hydrogen gas.

⇒ 2H⁺+ 2e^– = H₂ ↑ … … … … (2)

Now, equation (1)x2 + (2)x3,

2Al – 6e^– = 2Al³⁺

6H⁺+ 6e^– = 3H₂

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2Al + 6H⁺ = 2Al³⁺ + 3H₂

Or, adding necessary ions and radicals we get,

2Al + 3H₂SO₄ = Al₂(SO₄)₃ + 3H₂

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Al + H2SO4 = Al2(SO4)3 + SO2 + H2O