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Sulfuric acid dissolves aluminum metal according to the following reaction: 2Al(s)+3H2SO4(aq)→Al2(SO4)3(aq)+3H2(g)2Al(s)+3H2SO4(aq)→Al2(SO4)3(aq)+3H2(g) Suppose you wanted to dissolve an aluminum block with a mass of 15.4 gg .

What minimum mass of H2SO4 would you need?

What mass of H2 gas would be produced by the complete reaction of the aluminum block?

Answer :

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Answer:

83.9g of sulfuric acid is the minimum mass you would need

1.73g of hydrogen would be produced

Explanation:

Based on the reaction:

2 Al(s) + 3 H₂SO₄(aq) → Al₂(SO₄)₃(aq) + 3 H₂(g)

2 moles of solid aluminium react with 3 moles of sulfuric acid. Also, two moles of Al produce 3 moles of hydrogen gas.

15.4g of Al are:

15.4g Al × (1mol / 26.98g) = 0.571 moles of Al.

Moles of sulfuric acid:

0.571 moles Al × (3 mol H₂SO₄ / 2 mol Al) = 0.8565 moles H₂SO₄

In grams:

0.8565 moles H₂SO₄ × (98g / 1mol) = 83.9g of sulfuric acid is the minimum mass you would need

In the same way, moles of hydrogen produced are:

0.571 moles Al × (3 mol H₂ / 2 mol Al) = 0.8565 moles H₂

In grams:

0.8565 moles H₂ × (2.015g / 1mol) = 1.73g of hydrogen would be produced

The minimum mass of H₂SO₄ that would be needed is 84.0 g

The mass of H₂ gas that would be produced is 1.7 g

From the given balanced chemical equation for the reaction,  

2Al(s) + 3H₂SO₄(aq) → Al₂(SO₄)₃(aq) + 3H₂(g)

This means

3 moles of sulfuric acid is required to dissolve 2 moles of aluminum

Now, we will determine the number of moles of Aluminum present

Mass of aluminum present = 15.4 g

From the formula,  

[tex]Number\ of\ moles= \frac{Mass}{Atomic\ mass}[/tex]

Atomic mass of aluminum = 26.98 g/mol

∴ Number of moles of Al present = [tex]\frac{15.4}{26.98}[/tex]  

Number of moles of Al present = 0.5708 mole

Now,

Since 3 moles of sulfuric acid is required to dissolve 2 moles of aluminum

Then,  

x moles of sulfuric acid will be required to dissolve 0.5708 mole of aluminum

x = [tex]\frac{3\times 0.5708}{2}[/tex]

x = 0.8562 mole

∴The number of moles of sulfuric acid required to dissolve the aluminum is 0.8562 mole

Now, for the mass of sulfuric acid (H₂SO₄) that would be needed

Using the formula  

Mass = Number of moles × Molar mass  

Molar mass of H₂SO₄ = 98.079 g/mol

∴  The minimum mass of H₂SO₄ that would be needed = 0.8562 × 98.079

Minimum mass of H₂SO₄ that would be needed = 83.975 g

Minimum mass of H₂SO₄ that would be needed ≅ 84.0 g

Hence, the minimum mass of H₂SO₄ that would be needed is 84.0 g

For the mass of H₂ gas that would be produced

From the balanced chemical equation

3 moles of sulfuric acid dissolve 2 moles of aluminum to produce 3 moles of H₂ gas

Then,

0.8562 mole of sulfuric acid will dissolve 0.5708 mole of aluminum to produce 0.8562 mole of H₂ gas

∴ Number of moles of H₂ gas that would be produced is 0.8562 mole

Now, for the mass of H₂ gas that would be produced,

From the formula

Mass = Number of moles × Molar mass

Molar mass of H₂ = 2.016 g/mole

∴ Mass of H₂ gas that would be produced = 0.8562 × 2.016

Mass of H₂ gas that would be produced = 1.726 g

Mass of H₂ gas that would be produced ≅ 1.7 g

Hence, the mass of H₂ gas that would be produced is 1.7 g

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