Question

For the reaction 2 Al + 3 H2SO4⟶ 3 H2 + H2(SO4)3 how many grams of sulfuric acid, H2SO4, are needed to react completely with 53.7 g of aluminum, Al??

Answer 1

Answer:

292.4 g

Explanation:

• 2Al + 3H₂SO₄ → 3H₂ + Al₂(SO₄)₃

First we convert 53.7 g of Al into moles, using its molar mass:

• 53.7 g ÷ 27 g/mol = 1.99 mol Al

Then we convert Al moles into H₂SO₄ moles, using the stoichiometric coefficients:

• 1.99 mol Al * $\frac{3molH_2SO_4}{2molAl}$ = 2.98 mol H₂SO₄

Finally we convert 2.98 moles of sulfuric acid into grams, using its molar mass:

• 2.98 mol H₂SO₄ * 98 g/mol = 292.4 g