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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?

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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 Hâ‚‚SOâ‚„ that would be needed is 84.0 g

The mass of Hâ‚‚ 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 (Hâ‚‚SOâ‚„) that would be needed

Using the formula  

Mass = Number of moles × Molar mass  

Molar mass of Hâ‚‚SOâ‚„ = 98.079 g/mol

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

Minimum mass of Hâ‚‚SOâ‚„ that would be needed = 83.975 g

Minimum mass of H₂SO₄ that would be needed ≅ 84.0 g

Hence, the minimum mass of Hâ‚‚SOâ‚„ that would be needed is 84.0 g

For the mass of Hâ‚‚ 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 Hâ‚‚ gas

Then,

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

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

Now, for the mass of Hâ‚‚ gas that would be produced,

From the formula

Mass = Number of moles × Molar mass

Molar mass of Hâ‚‚ = 2.016 g/mole

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

Mass of Hâ‚‚ gas that would be produced = 1.726 g

Mass of H₂ gas that would be produced ≅ 1.7 g

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

Learn more here: https://brainly.com/question/10213672

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