Question

Sulfuric acid dissolves aluminum metal according to the following reaction:

2Al (s) + 3H₂SO₄ (aq) --> Al₂(SO₄)₃ (aq) + 3H₂ (g)
Suppose you wanted to dissolve an aluminum block with a mass of 15.0 g.

What minimum mass of H₂SO₄ (in g) would you need?

What mass of H₂ gas (in g) would be produced by the complete reaction of the aluminum block?

Answer 1

The minimum mass of H2SO4 needed to dissolve the aluminum block is 98 g, and the mass of H2 gas produced is 4/3 g.

Step-by-step explanation:

To find the minimum mass of $ ext{H}_2 ext{SO}_4$ needed, we need to determine the mole ratio between aluminum and sulfuric acid. From the balanced equation, we can see that 2 moles of aluminum reacts with 3 moles of sulfuric acid. The molar mass of sulfuric acid is 98 g/mol. So, to find the minimum mass of sulfuric acid, we can use the formula:

$ ext{mass of } ext{H}_2 ext{SO}_4 = ext{mole ratio} imes ext{molar mass of } ext{H}_2 ext{SO}_4$

Mass of $ ext{H}_2 ext{SO}_4 = rac{2}{2} imes rac{98 ext{ g}}{1 ext{ mole}} = 98 ext{ g}$

To find the mass of $ ext{H}_2$ gas produced, we can use the mole ratio between aluminum and hydrogen gas. From the balanced equation, we can see that 2 moles of aluminum produces 3 moles of hydrogen gas. The molar mass of hydrogen gas is 2 g/mol. So, the mass of $ ext{H}_2$ gas produced can be calculated as:

Mass of $ ext{H}_2 = ext{mole ratio} imes ext{molar mass of } ext{H}_2$

Mass of $ ext{H}_2 = rac{2}{3} imes rac{2 ext{ g}}{1 ext{ mole}} = rac{4}{3} ext{ g}$