Consider the reaction: 2 Al (s) + 3 H2SO4 (aq) ----Al2(SO4)3 (aq) + 3 H2 (g)

Question

What mass (in micrograms) of aluminum would completely react with 26.5 mL of 0.542 M sulfuric acid?

Answer 1

2Al(s) + 3H2SO4(aq) ==> Al2(SO4)3(aq) + 3H2(g) ... balanced equation

Using the stoichiometry of the balanced equation (2 mol Al : 3 mol H₂SO₄) and dimensional analysis, we can find the mass of Al needed to react with 26.5 ml of 0.542 M H2SO4.

How many moles H2SO4 are present?

26.5 ml x 1 L/1000 ml x 0.542 mol/L = 0.014363 moles H₂SO₄

How many moles Al are needed to react with this?

0.014363 mol H₂SO₄ x 2 mol Al / 3 mol H₂SO₄ = 0.009575 moles Al

Converting this to grams and then to micrograms (ug), we have...

0.009575 mol Al x 26.98 g Al/mol x 10⁶ ug//g = 2.58x10⁵ ug Al needed