Sulfuric acid can dissolve 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 24.4g .
What minimum amount of H2SO4 in grams would you need?
You have the balanced equation. That's always the first step.
Convert 24.4 g Al to mols. mols = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols Al to mols H2SO4.
Now convert mols H2SO4 to grams. g = mols x molar mass = ?
To determine the minimum amount of H2SO4 needed to dissolve the aluminum block, you need to consider the stoichiometry of the balanced chemical equation, balancing the number of moles of reactants and products.
According to the balanced chemical equation:
2Al(s) + 3H2SO4(aq) → Al2(SO4)3(aq) + 3H2(g)
The stoichiometric ratio between aluminum (Al) and sulfuric acid (H2SO4) is 2:3. This means that for every 2 moles of aluminum, we need 3 moles of sulfuric acid.
First, we need to convert the mass of aluminum (24.4g) to moles:
Mass of Al = 24.4g
Molar mass of Al = 26.98 g/mol (atomic mass of Al)
Number of moles of Al = Mass of Al / Molar mass of Al
Number of moles of Al = 24.4g / 26.98 g/mol ≈ 0.904 mol
Since the stoichiometric ratio between Al and H2SO4 is 2:3, we can determine the moles of H2SO4 required using the ratio:
Number of moles of H2SO4 = (3/2) * Number of moles of Al
Number of moles of H2SO4 = (3/2) * 0.904 mol ≈ 1.356 mol
Finally, we need to convert the moles of H2SO4 to grams using its molar mass:
Molar mass of H2SO4 = 98.09 g/mol (atomic mass of H2SO4)
Mass of H2SO4 = Number of moles of H2SO4 * Molar mass of H2SO4
Mass of H2SO4 = 1.356 mol * 98.09 g/mol ≈ 133.19 g
Therefore, the minimum amount of H2SO4 needed to dissolve the aluminum block with a mass of 24.4g is approximately 133.19 grams.