Sulfuric acid dissolves aluminum metal according to the above reaction. What mass (in g) of H2SO4(aq)would be needed to completely dissolve a12.7 g block of al?

Question

How Would I Solve this problem step by step to get the mass? 2 Al(s)+3H2SO4(aq) yield to Al2(SO4)3(aq)+3H2(g)

Answer 1

2Al (s) + 3H₂SO₄ (aq) → Al₂(SO₄)₃ (aq) + 3H₂ (g)

 

• MM_Al = 26.982 g/mol
• MM_H2SO4 = 98.079 g/mol

 

12.7 g Al x (1 mol Al / 26.982 g Al) x (3 mol H₂SO₄ / 2 mol Al) x (98.079 g H₂SO₄ / 1 mol H₂SO₄)

 

Ans. 69.2 g H₂SO₄